For $f \in \mathfrak{F}$ and $z \in \Bbb D$ the AM-GM inequality gives
$$
|f(z)| \le \sum_{n=0}^\infty |a_n z^n| \le
\sum_{n=0}^\infty \frac 12 \bigl( |a_n|^2 + |z|^{2n} \bigr)
= \frac 12 \bigl( \sum_{n=0}^\infty |a_n|^2
+ \sum_{n=0}^\infty |z|^{2n} \bigr) \le
\frac 12 \bigl( 1 + \frac{1}{1-|z|^2} \bigr)
$$
so that $\mathfrak{F}$ is uniformly bounded on compact subsets
of $\Bbb D$. Alternatively, one can use the Cauchy-Schwarz inequality for
infinite series to conclude that
$$
|f(z)| \le \frac{1}{\sqrt{1-|z|^2}} \, .
$$
It follows from Montel's theorem that
$\mathfrak{F}$ is normal, i.e. every sequence in $\mathfrak{F}$ has a subsequence which converges uniformly on
compact subsets of $\Bbb D$.
Now let $(f_k), f_k \in \mathfrak{F}$ be a sequence such that
$$
\lim_{k \to \infty} \text{Re} \, f_k(0) = \sup\{\text{Re} \, f(0) : f\in\mathfrak F\}
$$
$(f_k)$ has a subsequence $(f_{k_j})$ which converges uniformly on compact
subsets to a function $g$ which is holomorphic in $\Bbb D$.
It is clear that $g(\frac 12) = 0$ and
$$
\text{Re} \, g(0) = \sup\{\text{Re} \, f(0) : f\in\mathfrak F\} \, .
$$
Finally, since $\lim_{j \to \infty} f_{k_j}^{(n)}(0) = g^{(n)}(0)$ for all $n$,
$$
\sum_{n=0}^\infty \left( \frac{|g^{(n)}(0)|}{n!}\right)^2\leq 1
$$
so that $g \in \mathfrak F$. (Hint: consider a finite sum
$\sum_{n=0}^N$ first.)
(In response to your comment:) The extremal function $g$
satisfies $g(0) > 0$.
To see this, consider the function $\tilde g$ defined by
$$
\tilde g(z) = \frac{|g(0)|}{g(0)} g(z) \, .
$$
Then $\tilde g \in \mathfrak F$, so that
$$
\text{Re} \, g(0) \ge \text{Re} \, \tilde g(0) = \text{Re} \, |g(0)| = |g(0)| \ge \text{Re} \, g(0) .
$$
So equality must host in this chain:
$$
\text{Re} \, g(0) = |g(0)|
$$
which implies that $g(0)$ is a positive real number.
Remark: $\sum |a_n|^2\leq 1$ does not imply $f:\mathbb{D}\rightarrow\mathbb{D}$. For example,
$$
f(z) = - \frac{\sqrt 6}{\pi} \log(1-z) = \sum_{n=0}^\infty \frac{\sqrt 6}{\pi n} z^n
$$ is not bounded in $\Bbb D$, but has
$$
\sum_{n=0}^\infty |a_n|^2 = \frac {6}{\pi^2 }\sum_{n=0}^\infty \frac{1}{n^2} = 1 \, .
$$
