Let $L(x_0)$ denote the set of points $y\in p^{-1}(x_0)$ such that every loop $f$ with basepoint $x_0$ lifts to a loop $\tilde{f}$ with basepoint $y$. Note that the latter means that the unique lift $\tilde{f}$ of $f$ with $\tilde{f}(0) = y$ has the property $\tilde{f}(1) = y$.
Let $S$ denote the set of sections $s : X \to \tilde X$ and define $\phi : S \to p^{-1}(x_0), \phi(s) = s(x_0)$.
$\phi$ is injective: Each section $s$ is a lift of $id_X$, thus $s$ is uniquely determined by the value $s(x_0)$. Here we use the fact that $X$ is connected.
$\phi(S) \subset L(x_0)$: For any loop $f$ with basepoint $x_0$ the loop $s \circ f$ is a lift of $f$ with basepoint $y$.
It remains to show that each $y \in L(x_0)$ is in the image of $\phi$. We need some preprations.
For each path component $C$ of $\tilde X$, the restriction $p\mid_C : C \to X$ is a covering map.
i) $p \mid_C$ is surjective: Let $x \in X$. Pick $c_0 \in C$ and choose a path $u : I \to X$ such that $u(0) = p(c_0)$ and $u(1) = x$. There exists a lift $\tilde u$ of $u$ such that $\tilde u (0) = c_0$. Then $\tilde u(1) \in C$ and $p(\tilde u(1)) = u(1) = x$ which shows that $x \in p(C)$.
ii) Let $x \in X$. Choose a path connected open neigborhood $U$ of $x$ which is evenly covered by $p$. Then $p^{-1}(U) = \bigcup_{\alpha \in A} V_\alpha$ with pairwise disjoint open $V_\alpha \subset \tilde X$ which are mapped by $p$ homeomorphically onto $U$. Thus each $V_\alpha$ is path connected and therefore contained in some path component of $\tilde X$. Hence $(p \mid_C)^{-1}(U) = \bigcup_{\alpha \in A_C} V_\alpha$ for a suitable nonempty subset $A_C \subset A$. This means that $U$ is evenly covered by $p \mid_C$.
The cardinality of $(p \mid_C)^{-1}(x_0)$ is the number of sheets of $p \mid_C$. Let $y \in p^{-1}(x_0)$ and $C$ its path component. We claim that
(a) If $p \mid_C$ has one sheet, then $y \in \phi(S) \subset L(x_0)$.
(b) If $p \mid_C$ has more than one sheet, then $y \notin L(x_0)$.
This clearly proves $\phi(S) = L(x_0)$.
(a) In this case $p \mid_C$ is an open bijection, i.e. a homeomorphism. Thus $s = (p \mid_C)^{-1} : X \to C \hookrightarrow \tilde X$ is a section such that $s(x_0) = y$ which means $y \in \phi(S)$.
(b) In this case there exists a point $y' \in C$ other than $y$ such that $p(y') = x_0$. There is a path $\tilde f$ in $C$ from $y$ to $y'$; it is no loop. On the other hand f = $p \circ \tilde f$ is a loop based at $x_0$. But since $\tilde f$ is the unique lift of $f$ with $\tilde f(0) = y$, we see that $y \notin L(x_0)$.
Update:
Instead of steps 3. and 4. we can invoke the following two well-known theorems:
(1) Let $p:\tilde{X}\to X$ be a covering map, $x_0 \in X$ and $y \in p^{-1}(x_0)$. The induced homomorphism $p_* : \pi_1(\tilde X,y) \to \pi_1(X,x_0)$ is injective. The image subgroup $p_*(\pi_1(\tilde X,y)) \subset \pi_1(X,x_0)$ consists of the homotopy classes of loops in $X$ based at $x_0$ whose lifts to $\tilde X$ starting at $y$ are loops.
(2) Lifting theorem. Let $p:\tilde{X}\to X$ be a covering map, $x_0 \in X$ and $y \in p^{-1}(x_0)$ and $g : (Z, z_0) \to (X,x_0)$ be a map with $Z$ path-connected and locally path connected. Then a lift $\tilde g : (Z, z_0) \to (\tilde X, y)$ of $g$ exists if and only if $g_*(\pi_1((Y, y_0)) \subset p_*(\pi_1(\tilde X,y))$.
Now (1) says that $y \in L(x_0)$ if and only if $p_*(\pi_1(\tilde X,y)) = \pi_1(X,x_0)$ and (2) says that the identity $id_X : (X,x_0) \to (X,x_0)$ has a lift $s : (X,x_0) \to (\tilde X,y)$ (which is nothing else than a section of $p$ with $\phi(s) =y$) if and only if $\pi_1(X,x_0) = (id_X)_*(\pi_1((X, x_0)) \subset p_*(\pi_1(\tilde X,y))$. The latter condition means $\pi_1(X,x_0) = p_*(\pi_1(\tilde X,y))$. This shows $L(x_0) = \phi(S)$.
