holomorphic functions uniformly convergent on any compact subset of an open set implies uniformly convergent on the whole open set. In Stein's complex analysis, there is one theorem 5.3 of Chapter 2 which says that if holomorphic functions  $f_n$ converges uniformly to $f$ on any compact subset of an open set $\Omega$, then so does $f'_n$. 
In his proof, he said "without loss of generality we can assume $f_n$ converges uniformly to $f$ on all of $\Omega$". So I think what he meant was that from the conditions of the theorem, we can get $f_n$ converges uniformly to $f$ on all of $\Omega$.
However, I don't know how to prove this assumption. Any help would be welcome.
 A: Here's a useful topological fact. 
Fact. If $K$ is a compact subset of a domain $\Omega$, then there exists a domain $U$ such that $K\subset U$ and $\overline{U}$ is a compact  subset of $\Omega$. 
Proof: Fix a point $z_0\in \Omega$. For $n=1,2,\dots$ let $G_n$ be the connected component of $z_0$ in the set $$\{z\in\Omega: |z|<n, \ \operatorname{dist}(z,\partial \Omega)>1/n\}$$
Using the connectedness of $\Omega$, one can show that $\bigcup_n G_n=\Omega$. Therefore, the sets $G_n$ form an open cover of $K$. There is a finite subcover; since the sets $G_n$ are nested, this means $K\subset G_n$ for some $n$. This $G_n$ is the desired $U$. $\quad\Box$ 
Back to the issue. It's not really about holomorphic functions and their derivatives. We are to prove the following: 

Theorem 1. If statement $A$ holds on every compact subset of domain $\Omega$, then statement $B$ holds on every compact subset of $\Omega$.

But instead we prove   

Theorem 2. If statement $A$ holds on domain $\Omega$, then statement $B$ holds on every compact subset of $\Omega$.

Indeed, suppose Theorem 2 is proved. Given $\Omega$ as in Theorem 1 and its compact set $K$,  take $U$ from the topological fact. Since $\overline{U}$ is a compact subset of $\Omega$, property $A$ holds on $U$. Theorem 2 says that property $B$ holds on $K$, which was to be proved. 
A: Let us assume the following result: if $\{f_n\}$ is a sequence of functions on some open set $U$ which converge uniformly on $U$ to $f$ (i.e. $\sup_{U}|f_n-f|\to 0$ for $n\to\infty$), then the sequence $\{f'_n\}$ converges as well uniformly on the compact sets of $U$ to a function $g$ and we have $f'=g$. This is the theorem proved in the book by Stein, if I understand well.
Now, let $\{f_n:\Omega\to\mathbb{C}\}_n$ be sequence of holomorphic functions converging uniformly on compact subsets of $\Omega$ to $f$. For any compact set $K\subset\Omega$, we can find an open set $U$ such that $K\subset U\Subset \Omega$. 
As $U\Subset\Omega$, $f_n\vert_{U}\to f\vert_{U}$ uniformly: $F=\overline{U}$ is compact, hence
$$\sup_{U}|f_n\vert_U-f\vert_U|=\sup_{U}|f_n-f|=\sup_F|f_n-f|\to 0$$
when $n\to \infty$. Therefore, by our assumption $(f_n\vert_U)'$ converges uniformly on compact sets of $U$ to $g$, such that $g=f'$.
Obviousy, $(f_n\vert_U)'=f_n'\vert_U$ and then 
$$\sup_{K}|f'_n-f'|=\sup_K|f'_n\vert_U-f'\vert_U|\to 0$$ 
when $n\to\infty$. And this is the thesis.
