Even though OP did not asked for proofs for both statements, I cannot stop writing down two proofs, and providing them here for future reference.
So, the statement P1 goes as follows:
Let $(V,\|\ \cdot\ \|)$ be a real Banach space and $K\subset V$ be a (norm) closed convex set. Assume that $V$ is reflexive. Show that for all $x\in V$ there exists $y\in K$ such that $$\|x-y\|=\inf_{z\in K}\|x-z\|=\text{dist}(x, K).$$
Proof:
We firstly prove that $\varphi:V\longrightarrow(-\infty,\infty]$ defined by $\varphi(x):=\|x\|$ is a convex and continuous function. Let $\theta\in [0,1]$ and $x,y\in V$, then by the triangle inequality, we have $$\varphi(\theta x+(1-\theta)y)=\|\theta x+(1-\theta)y\|\leq \|\theta x\|+\|(1-\theta)y\|=\theta\|x\|+(1-\theta)\|y\|=\theta\varphi(x)+(1-\theta)\varphi(y).$$
Now, let $\epsilon>0$ and $(x_{n})$ be a sequence in $V$ that converges to $x\in V$. Then, by definition, there exists $N$ large enough such that for all $n>N$, we have $\|x_{n}-x\|<\epsilon$. On the other hand, by (inverse) triangle inequality $$\epsilon>\|x_{n}-x\|\geq \big|\ \|x_{n}\|-\|x\|\ \big|=|\varphi(x_{n})-\varphi(x)|.$$ This means that when $x_{n}\rightarrow x$ we have that $\varphi(x_{n})\rightarrow\varphi(x)$, as desired.
In particular, the norm is convex and (strong) lower semi-continuous.
Let $V$ be a reflexive real Banach space and $K\subset V$ a (norm) closed convex subset. Let $x\in V$, consider a sequence $(y_{n})_{n=1}^{\infty}$ such that $y_{n}\in K$ for all $n$ and $$\lim_{n\rightarrow\infty}\|x-y_{n}\|=\text{dist}(x,K)=\inf_{z\in K}\|x-z\|.$$
Note that such a sequence exists. Indeed, as the set $\{\|x-z\|:z\in K\}$ is bounded below by zero, the infimum $\inf_{z\in K}\|x-z\|$ exists. We denote $d$ to be this infimum. By definition, for each $n\in\mathbb{N}$, there exists $y_{n}\in K$ such that $$d\leq \|x-y_{n}\|\leq d+\dfrac{1}{n}.$$
Taking $n\rightarrow\infty$, we see that $\lim_{n\rightarrow\infty}\|x-y_{n}\|=d.$ Such a (minimizing) sequence $(y_{n})_{n=1}^{\infty}$ is exactly what we need. In particular, this sequence is (norm) bounded because $\text{dist}(x,K)<\infty$.
As $V$ is reflexive and $(y_{n})_{n=1}^{\infty}$ is a (strongly) bounded sequence, it follows from [Theorem 3.18, Haim] that there exists a subsequence $(y_{n_{k}})$ that converges in the weak topology $\sigma(V,V^{*})$.
Let $y$ be the weak limit of this subsequence. As $K$ is strongly closed (since it is norm closed) and $K$ is convex, it follows from [Theorem 3.7], Haim that $K$ is weakly closed.
Hence, $y\in K$.
As we have proved that the norm is convex and (strongly) lower semi-continuous, it follows from [Corollary 3.9, Haim] that the norm is weakly lower semi-continuous.
It follows that $$\|x-y\|\leq\liminf_{n_{k}\rightarrow\infty}\|x-y_{n_{k}}\|=\inf_{z\in K}\|x-z\|.$$ But $\inf_{z\in K}\|x-z\|\leq \|x-u\|$ for all $u\in K$, and thus in particular $\inf_{z\in K}\|x-z\|\leq \|x-y\|.$
Hence, we have $\|x-y\|=\inf_{z\in K}\|x-z\|$, which concludes the proof.
Now, let us recall the definition of uniform convexity:
$V$ is said to be uniformly convex if for all $\epsilon\in(0,2)$, there exists $\delta_{\epsilon}>0$ such that for all $x,y\in V$, one has $$\|x\|\leq 1,\|y\|\leq 1\ \text{and}\ \|x-y\|\geq\epsilon\implies \Bigg\|\dfrac{x+y}{2}\Bigg\|\leq 1-\delta_{\epsilon}.$$
Then, statement P2 is the following:
Assume that $V$ is uniformly convex, show that for all $x\in V$, there exists a unique $y\in K$ such that $$\|x-y\|=\inf_{z\in K}\|x-z\|=\text{dist}(x,K).$$
Proof:
To show the existence of such a unique element, the key is to show that a minimizing sequence is Cauchy.
Let $x\in V$, we denote $d:=\text{dist}(x,K)$. Then, we consider a minimizing sequence $(y_{n})_{n=1}^{\infty}$ whose existence has been proved in part $(i)$. For each $n$, we define $$x_{n}:=y_{n}-x, \ \ d_{n}:=\|y_{n}-x\|\ \ \text{and}\ \ z_{n}:=\dfrac{x_{n}}{d_{n}}.$$
It is clear that $z_{n}$ is a unit vector for each $n$ and $d_{n}\rightarrow d$ as $n\rightarrow\infty$. Also, we can write
\begin{align*}
\dfrac{z_{n}+z_{m}}{2}=\dfrac{x_{n}}{2d_{n}}+\dfrac{x_{m}}{2d_{m}}=\Bigg(\dfrac{1}{2d_{n}}+\dfrac{1}{2d_{m}}\Bigg)(&c_{n}x_{n}+c_{m}x_{m}),\\
&\text{where}\ \ c_{n}:=\dfrac{1}{2d_{n}(\frac{1}{2d_{n}}+\frac{1}{2d_{m}})}\ \ \text{and}\ \ c_{m}:=\dfrac{1}{2d_{m}(\frac{1}{2d_{n}}+\frac{1}{2d_{m}})}.
\end{align*}
Note that $c_{n}+c_{m}=1$. Indeed, $$c_{n}+c_{m}=\dfrac{1}{1+\frac{d_{n}}{d_{m}}}+\dfrac{1}{1+\frac{d_{m}}{d_{n}}}=\dfrac{1+\frac{d_{m}}{d_{n}}+1+\frac{d_{n}}{d_{m}}}{(1+\frac{d_{n}}{d_{m}})(1+\frac{d_{m}}{d_{n}})}=\dfrac{1+\frac{d_{m}}{d_{n}}+1+\frac{d_{n}}{d_{m}}}{1+\frac{d_{m}}{d_{n}}+\frac{d_{n}}{d_{m}}+1}=1.$$
Therefore, by convexity of $K$, we know that $(c_{n}y_{n}+c_{m}y_{m})\in K$. Hence, it follows that
\begin{align*}
\Bigg\|\dfrac{z_{n}+z_{m}}{2}\Bigg\|=\Bigg\|\dfrac{x_{n}}{2d_{n}}+\dfrac{x_{m}}{2d_{m}}\Bigg\|&=\Bigg(\dfrac{1}{2d_{n}}+\dfrac{1}{2d_{m}}\Bigg)\|c_{n}y_{n}+c_{m}y_{m}-c_{n}x-c_{m}x\|\\
&=\Bigg(\dfrac{1}{2d_{n}}+\dfrac{1}{2d_{m}}\Bigg)\|c_{n}y_{n}+c_{m}y_{m}-(c_{n}+c_{m})x\|\\
&=\Bigg(\dfrac{1}{2d_{n}}+\dfrac{1}{2d_{m}}\Bigg)\|c_{n}y_{n}+c_{m}y_{m}-x\|\\
&=\Bigg(\dfrac{1}{2d_{n}}+\dfrac{1}{2d_{m}}\Bigg)\|x-(c_{n}y_{n}+c_{m}y_{m})\|\\
&\geq \Bigg(\dfrac{1}{2d_{n}}+\dfrac{1}{2d_{m}}\Bigg)\inf_{z\in K}\|x-z\|\\
&=\dfrac{d}{2d_{n}}+\dfrac{d}{2d_{m}}.
\end{align*}
Now, let $\epsilon\in (0,2)$ and let $\delta_{\epsilon}$ be the correspondence in the definition of uniform convexity. Then, we can find a $\epsilon_{1}>0$ small enough such that $\frac{d}{d+\epsilon_{1}}>1-\delta_{\epsilon}$ and an $N$ large enough such that $d\leq d_{n}<d+\epsilon_{1}$ for all $n\geq N$. (Such an $N$ exists by definition of infimum and $d\leq d_{n}$ because $d$ is the infimum.)
We now show that $(z_{n})_{n=1}^{\infty}$ is Cauchy. To show this, we need to show that for all $n,m\geq N$, we have $\|z_{n}-z_{m}\|=\|\frac{x_{n}}{d_{n}}-\frac{x_{m}}{d_{m}}\|<\epsilon.$ Suppose not, then there exists $n,m\geq N$ such that $$\|z_{n}-z_{m}\|=\Bigg\|\dfrac{x_{n}}{d_{n}}-\dfrac{x_{m}}{d_{m}}\Bigg\|\geq\epsilon.$$ But then, $$\Bigg\|\dfrac{z_{n}+z_{m}}{2}\Bigg\|\geq \dfrac{1}{2}\Bigg(\dfrac{d}{d_{n}}+\dfrac{d}{d_{m}}\Bigg)>\dfrac{1}{2}\Bigg(\dfrac{d}{d+\epsilon_{1}}+\dfrac{d}{d+\epsilon_{1}}\Bigg)>\dfrac{1}{2}\cdot(1-\delta_{\epsilon}+1-\delta_{\epsilon})=1-\delta_{\epsilon}.$$
To summarize, we have $\|z_{n}\|=1=\|z_{m}\|$ with $\|z_{n}-z_{m}\|\geq\epsilon$, but then $\|\frac{z_{n}+z_{m}}{2}\|\geq 1-\delta_{\epsilon}.$ This contradicts the uniform convexity, and thus $(z_{n})_{n=1}^{\infty}$ must be Cauchy.
(Just a side note here that $N$ indeed depends on $\epsilon$, so the Cauchy argument makes sense. It depends on $\epsilon$, because we choose $N$ that depends on $\epsilon_{1}$, but $\epsilon_{1}$ depends on $\delta_{\epsilon}$ which then depends on $\epsilon$, so we are good.)
As $(z_{n})$ is Cauchy and $d_{n}, d_{m}\rightarrow d$, we have $$0=\lim_{n,m\rightarrow\infty}\|z_{n}-z_{m}\|=\lim_{n,m\rightarrow\infty}\Bigg\|\dfrac{x_{n}}{d_{n}}-\dfrac{x_{m}}{d_{m}}\Bigg\|=\lim_{n,m\rightarrow\infty}\Bigg\|\dfrac{y_{n}-x-y_{m}+x}{d}\Bigg\|,$$ which implies that $$\lim_{n,m\rightarrow\infty}\|y_{n}-y_{m}\|=0.$$
This shows that the minimizing sequence $(y_{n})$ is Cauchy. As $V$ is complete and $K$ is closed, $y_{n}$ converges to an element $y\in K$.
As norm is continuous, we have $\|x-y_{n}\|\rightarrow \|x-y\|$ but as $(y_{n})$ is a minimizing sequence, by definition we have $\|x-y_{n}\|\rightarrow \text{dist}(x,K).$
Therefore, we proved the existence of $y\in K$ such that $\|x-y\|=\text{dist}(x,k).$
To show the uniqueness of such $y$, suppose there exist $y\neq y'\in K$ such that $$\|x-y\|=\|x-y'\|=\text{dist}(x,K):=d$$ Then, as $y\neq y'$, we must have $$\Bigg\|\dfrac{x-y}{d}-\dfrac{x-y'}{d}\Bigg\|=\dfrac{1}{d}\|y-y'\|\geq\epsilon>0,\ \ \text{for some (actually any)}\ \ \epsilon\in (0,2).$$
As the space is uniformly convex and $\|\frac{x-y}{d}\|=1=\|\frac{x-y'}{d}\|$, there exists $\delta_{\epsilon}>0$ such that $$\dfrac{1}{2}\Bigg\|\dfrac{x-y}{d}+\dfrac{x-y'}{d}\Bigg\|=\dfrac{1}{d}\Bigg\|x-\dfrac{y+y'}{2}\Bigg\|\leq 1-\delta_{\epsilon}<1.$$ This implies that $$\Bigg\|x-\dfrac{y+y'}{2}\Bigg\|<d.$$
But $K$ is convex and $y,y'\in K$ and thus $\frac{y+y'}{2}=\frac{1}{2}y+(1-\frac{1}{2})y'\in K$ as well. Then the above strict inequality gives us a contradiction because $d$ is the infimum of $\|x-z\|$ where $z$ is over all elements of $K$.
Hence, $y$ must be unique. The proof is concluded.
