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. 

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.  
