Here is my answer following the prompts in the comments.
Let $z_0 \in \Omega, \epsilon > 0$ be given and choose $R > 0$ such that $\overline{D_{R + \epsilon}(z_0)} \subset \Omega$. Then, there exists $N \in \mathbb{N}$ so that
$$n \geq N \implies \int_{D_{R + \epsilon}(z_0)}|f_n(x+iy) - f(x+iy)|\, dx \, dy < \frac{|D_{\epsilon}(z_0)|\epsilon}{2}.$$
and in particular
$$n,m \geq N \implies \int_{D_{R + \epsilon}(z_0)} |f_n(x+iy) - f_m(x+iy)|\, dx \, dy < |D_{\epsilon}(z_0)|\epsilon.$$ Also given $f_n, f_m$ are holomorphic in $\Omega$ then by mean-value property
$$f_n(z_0) = \frac{1}{2\pi} f_n(z_0 + r e^{i \theta})\, d \theta, \quad f_m(z_0) =f_m(z_0 + r e^{i \theta})\, d \theta$$ for all $0 < r \leq R$. Then by the use of polar coordinates whenever $n, m \geq N$
\begin{align}
|D_R(z_0)| |f_n(z_0) - f_m(z_0)| &= \pi R^2 |f_n(z_0) - f_m(z_0)|\\
&= \left| \int_{D_R(z_0)}[f_n(x+iy) - f_m(x+iy)] \, dx \, dy \right| \\
&\leq \int_{D_{R + \epsilon}(z_0)}\left| f_n(x+iy) - f_m(x+iy) \right|\, dx \, dy \\
&\leq |D_\epsilon(z_0)| \epsilon
\end{align} Thus, $n,m \geq N \implies |f_n(z_0) - f_m(z_0)| \leq \frac{|D_\epsilon(z_0)|}{|D_R(z_0)|}\epsilon < \epsilon.$
Given any $z_1 \in \overline{D_R(z_0)}$ then $\overline{D_\epsilon(z_1)} \subset \overline{D_{R + \epsilon}(z_0)}$ so
\begin{align}
|D_\epsilon(z_1)| |f_n(z_1) - f_m(z_1)| &= \pi \epsilon^2 |f_n(z_1) - f_m(z_1)|\\
&= \left| \int_{D_\epsilon(z_1)}[f_n(x+iy) - f_m(x+iy)] \, dx \, dy \right| \\
&\leq \int_{D_{R + \epsilon}(z_0)}\left| f_n(x+iy) - f_m(x+iy) \right|\, dx \, dy \\
&\leq |D_\epsilon(z_0)| \epsilon
\end{align} Thus, $n,m \geq N \implies |f_n(z_1) - f_m(z_1)| \leq \epsilon.$ Therefore, $\{f_n\}$ is uniformly Cauchy in $\overline{D_R(z_0)}$ so $\{f_n\}$ is uniformly convergent in $\overline{D_R(z_0)}$. As such closed disks are arbitrary we have $f_n$ converges uniformly on every compact subsets of $\Omega$ implying that $f_n$ converges uniformly to $F$, a holomorphic function in $\Omega$. Now, we are left to show that $F(z) = f(z), \forall z \in \Omega.$ Given any closed disk $D \subset \Omega$, we have
$$0 = \lim_{n \to \infty} \int_D |f_n(x+iy) - f(x+iy)|\, dx \, dy = \int_D |F(x+iy) - f(x+iy)|\, dx \, dy$$ thus $F = f$ a.e. in $D$. If for a contradiction that the set $\{z \in D \colon f(z) \neq F(z)\} = (f - F)^{-1}(D - \{0\})$ is nonempty then it being open must have positive measure, contrary to our conclusion. Thus, $f(z) = F(z), \forall z \in D$. Again, $D$ is an arbitrary closed disk in $\Omega$ then $f(z) = F(z), \forall z \in \Omega$, this concludes the proof. As for the convergence of the derivatives, it follows a standard argument available in most complex analysis text.
Please do let me know if there any gaps in my proof. Thank you!