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?

you basically want to show $\forall K\subseteq \Omega$ compact, $\forall \varepsilon > 0$ $\exists n$ such that $\forall x\in K,\forall m>n$ u have:
$|f_n(x)-f_m(x)|<\varepsilon$
So what you want to do is estimate this difference using the cauchy formula. The difference is a holomorphic function:
$|f_n(x)-f_m(x)|=|\sum_{k=0}^{\infty}\frac{1}{2\pi i}\oint \frac{f_n(z)-f_m(z)}{(z-a)^{k+1}}dz\cdot (x-a)^k|$
By dominated convergence you know that the integral will converge to zero, therefore you can make it arbitrarily small by choosing n big enough. But here comes the kicker: you have to do this such that the sum also converges ...Take an open covering of K with balls of radius $<1$ (this will make the sum summable) and select a finite subcovering (K is compact) then you obtain a $n$ for every ball $B$ in your covering such that 
$|f_n(x)-f_m(x)|<\varepsilon\quad \forall x\in B$
take the biggest $n$ of your covering and you are done.
