Uniform convergence of uniformly bounded sequence of holomorphic functions $\{f_n\}$ is a uniformly bounded sequence of holomorphic functions in $\Omega$ which pointwise converges. Prove that the convergence is uniform on every compact subset of $\Omega$. (Hint. Apply LDCT to the Cauchy formula for $f_n-f_m$.)
It is well known - I found one solution here.
However, in fact it is from Rudin RCA chapter 10, and Arzela-Ascoli is in the next chapter. So I believe that there is an relatively elementary proof. Anyone know about this?

Take any $\overline{B(z_0,r)} \subset D$, then $f_n$ is uniformly bounded on $\overline{B(z_0,r)}$. Hence $f_n'$ is uniformly bounded, say $|f_n'| \le M, z \in \overline{B(z_0,r)}$.
For any $\epsilon >0$, take $\delta=\frac{\epsilon}{3M},$ for $|z_1-z_2|<\delta$ implies that $$|f(z_1)-f(z_2)| =|\int_{z_2}^{z_1}f_n'(z)dz|< \frac{\epsilon}{3}$$
Hence $f_n$ equicontinuous on $B(z_0,r)$.
Take any compace subset $K \subset \Omega$, we have a finite cover $$K \subset \bigcup_{j=1}^l B(z_j,\frac{\delta}{2})$$
So $f_n$ also equicontinuous on $K$.
Since $$\lim_{n \to \infty}f_n(z_j)$$ exists, then for any $\epsilon >0$, there exists a $N_j \in \mathbb{Z_+}$, for $n,m>N_j$ implies that $|f_n(z_j)-f_m(z_j)| < \frac{\epsilon}{3}$.
Pick $N=max(N_1,……,N_j)$, for $n,m>N$, we have $$|f_n(z)-f_m(z)|<|f_n(z)-f_n(z_k)|+|f_n(z_k)-f_m(z_k)|+|f_m(z_k)-f_m(z)|<\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon$$
Here $z \in B(z_k, \frac{\delta}{2})(1 \le k \le l)$, and so $f_n$ uniformly converges on any compact subset of $\Omega$.
