Let $d > 0$. If $g$ is holomorphic on some neighbourhood of $\overline{D_d(z)}$, the mean-value theorem tells us that
$$g(z) = \frac{1}{2\pi} \int_0^{2\pi} g(z + \rho e^{i\varphi})\,d\varphi$$
for $0 \leqslant \rho \leqslant d$, hence
$$| g(z)| \leqslant \frac{1}{2\pi} \int_0^{2\pi} | g(z + \rho e^{i\varphi}|\,d\varphi. \tag{1}$$
Multiplying $(1)$ with $\rho$ and integrating from $0$ to $d$ we obtain
$$|g(z)| \frac{d^2}{2} \le \frac{1}{2\pi} \int_0^d \int_0^{2\pi} | g(z + \rho e^{i\varphi})|\,\rho\,d\varphi\,d\rho = \frac{1}{2\pi} \iint_{\lvert w - z\rvert \leqslant d} | g(w)|\,dx\,dy.\tag{2}$$
Letting $g = f^2$ in $(2)$ and dividing by $d^2/2$, then taking the square root, we obtain
$$|f(z)|\le \frac{1}{\sqrt{\pi}\,d} \,\lVert f\rVert_{L^2}(D_d(z)). \tag{3}$$
Choose $d = r-s$ to see
$$\lvert f(z)\rvert \le \frac{1}{\sqrt{\pi}\,(r-s)} \,\lVert f\rVert_{L^2(D_{r-s}(z))} \le \frac{1}{\sqrt{\pi}\,(r-s)} \,\lVert f\rVert_{L^2(D_r(z_0))}\tag{4}$$
for all $z \in D_s(z_0)$, whence
$$\lVert f\rVert_{L^{\infty}(D_s(z_0))} \le \frac{1}{\sqrt{\pi}\,(r-s)} \,\lVert f\rVert_{L^2(D_r(z_0))}.$$
If $0 < R < 1$ and $z \in \Bbb D$ with $|z| < 1-R$ then $\overline{D_R(z)} \subset \Bbb D$ and
$$
|f(z)| \le \frac{1}{2\pi}\int_0^{2\pi}|f(z+re^{i\theta})|\, d\theta
$$
for $0 \le r \le R$. Multiplying this inequality with $r$ and integration with respect to $r \in [0, R]$ gives
$$
\frac 12 R^2 |f(z)| \le \frac{1}{2\pi}\iint_{\overline{D_R(z)}} |f(x+iy)|^2dxdy \, ,
$$
i.e.
$$
|f(z)| \le \frac{1}{\pi R^2}\iint_{\overline{D_R(z)}} |f(x+iy)|^2dxdy \le \frac{1}{\pi R^2} \, .
$$
It follows that $\mathcal F$ is uniformly bounded on each disk $|z| < 1-R$, and therefore locally uniformly bounded.
Best Answer
Let $d > 0$. If $g$ is holomorphic on some neighbourhood of $\overline{D_d(z)}$, the mean-value theorem tells us that
$$g(z) = \frac{1}{2\pi} \int_0^{2\pi} g(z + \rho e^{i\varphi})\,d\varphi$$
for $0 \leqslant \rho \leqslant d$, hence
$$\lvert g(z)\rvert \leqslant \frac{1}{2\pi} \int_0^{2\pi} \lvert g(z + \rho e^{i\varphi})\rvert\,d\varphi. \tag{1}$$
Multiplying $(1)$ with $\rho$ and integrating from $0$ to $d$ we obtain
$$\lvert g(z)\rvert \frac{d^2}{2} \leqslant \frac{1}{2\pi} \int_0^d \int_0^{2\pi} \lvert g(z + \rho e^{i\varphi})\rvert\,\rho\,d\varphi\,d\rho = \frac{1}{2\pi} \iint_{\lvert w - z\rvert \leqslant d} \lvert g(w)\rvert\,dx\,dy.\tag{2}$$
Letting $g = f^2$ in $(2)$ and dividing by $d^2/2$, then taking the square root, we obtain
$$\lvert f(z)\rvert \leqslant \frac{1}{\sqrt{\pi}\,d} \,\lVert f\rVert_{L^2}(D_d(z)). \tag{3}$$
Choose $d = r-s$ to see
$$\lvert f(z)\rvert \leqslant \frac{1}{\sqrt{\pi}\,(r-s)} \,\lVert f\rVert_{L^2(D_{r-s}(z))} \leqslant \frac{1}{\sqrt{\pi}\,(r-s)} \,\lVert f\rVert_{L^2(D_r(z_0))}\tag{4}$$
for all $z \in D_s(z_0)$, whence
$$\lVert f\rVert_{L^{\infty}(D_s(z_0))} \leqslant \frac{1}{\sqrt{\pi}\,(r-s)} \,\lVert f\rVert_{L^2(D_r(z_0))}.$$
If $K \subset U$ is compact, choose a positive $d$ smaller than the distance from $K$ to $\mathbb{C}\setminus U$, so that $\overline{D_d(z)} \subset U$ for all $z\in K$. Estimating the right hand side of $(3)$ by $\frac{1}{\sqrt{\pi}\,d}\, \lVert f\rVert_{L^2(U)}$, we see
$$\lVert f\rVert_{L^{\infty}(K)} \leqslant \frac{1}{\sqrt{\pi}\,d} \,\lVert f\rVert_{L^2(U)}$$
for every square-integrable holomorphic $f \colon U \to \mathbb{C}$.
Thus if $(f_n)$ is an $L^2(U)$-Cauchy sequence of holomorphic functions, the restricted sequence $\bigl(f_n\rvert_K\bigr)$ is an $L^{\infty}(K)$-Cauchy sequence, hence $\bigl(f_n\rvert_K\bigr)$ converges uniformly. Since the compact subsets of $U$ cover $U$, it follows that $(f_n)$ converges pointwise on all of $U$, and by the above the convergence is uniform on all compact subsets of $U$. By Weierstraß' convergence theorem, the limit function is holomorphic on $U$.