Bounded functions and Montel's Theorem Let $\Omega\subset\mathbb{C}$ be a region and $(z_n)_n\subset\Omega$ such that $z_n\rightarrow z\in\Omega$. Also, let $f_n:\Omega\rightarrow\mathbb{C}$ be a sequence of $f_n\in\mathcal{H}(\Omega)$ such that there exists $M$ satisfying that for each $n\in\mathbb{N}$, $|f_n|<M$ and $\lim_{n\rightarrow\infty}|f_n(z_n)|=M$.
I have to show that $|f_n|\rightarrow M$ uniformly on the compact sets of $\Omega$.
By now I have noticed that $\mathcal{F}:=(f_n)_n$ is a normal family thanks to Montel's theorem, and also that 
$$|f_n(z_n)|<M\qquad\forall n\in\mathbb{N}.$$ 
How can I proceed with the rest of the proof? I have tried to prove it via the definition of each property, but I don't get anywhere. 
 A: Let $K \subset \Omega$ be a compact. Without LOG, we can suppose that $K$ contains an open disk centered on $x$.
Now let's proceed by contradiction, i.e. that $\vert f_n\vert $ doesn't converge uniformly to $M$. This means that it exists a real $0 < M^\prime < M$ and a strictly increasing sequence of integers $(n_j)_j$ and $z_j\in K$ with $\vert f_{n_j}(z_j)\vert \le M^\prime$. The sequence $(z_j)$ is included in the compact $K$ and has therefore a converging sub-sequence to a complex $z \in K$. Again without LOG, we can suppose that the sequence $(f_n)$ is such that there exists a sequence $(z_n) \to z$ with $\vert f_n(z_n) \vert \le M^\prime$.
Now using Montel's theorem, it exists a sub-sequence $(f_{n_k})$ converging uniformly to $f$ holomorphic in $K$. $f$ is not constant as
$$\vert f(z) \vert \le M^\prime < M=\vert f(x) \vert.$$
But this is contradicting the maximum modulus principle as $x$ is an interior point of $K$.
A: Hint. It suffices to show that every subsequence of $\{\,f_n\}$ possesses a sub-subsequence $f_{k_n}$, such that $|\,f_{k_n}|\to M$, uniformly in compact subsets.
As $\{f_n\}$ is bounded, then it possesses a subsequence $\{f_{k_n}\}$, uniformly convergent in compact subsets, and $f_{k_n}\to f$, and $f$ is analytic in $\Omega$ and $|\,f|\le M$. In fact, $|f_{k_n}(z_{k_n})|\to M$, and as $|f_{k_n}(z_{k_n})|\to |\,f(z)|=M$, then $f$ is constant, as it achieved an absolute maximum in an interior point and $\Omega$ is connected.  
