Given:
- a continuous function $f:\Omega\to\Bbb{C}$ $(\Omega\subset\Bbb{C})$
- $f = \lim\limits_{n\to\infty} f_n$ uniformly on compact subsets of $\Omega$
- a compact set $K\subset\Omega$
- $a\in\Omega\setminus K$
- $|f(a)| > \sup\limits_{K} |f|$
Then there exists $N\in\Bbb{N}$ such that $|f_N(a)| > \sup\limits_{K} |f_N|$
We can start be choosing a compact set $K_1$ containing $K$ and the point $a$. Now we have uniform approximation of $f$ on $K_1$ by $f_n$'s. Now to arrive at a contradiction we assume that no such $N$ exist. It is now possible to choose $N_1$ such that the inequality does not hold for $n>N_1$. Now I want to take the limit and arrive at a contradiction but I am not sure how to do that.
Am I in the right direction? Is there any other way to prove this which is easy to see?
Taking a limit of a statement like this is going to be tricky because of the $\sup_K$. Instead of trying to make sense of that, I would suggest directly using the definition of uniform convergence. For any $\epsilon>0$, you can find $N$ such that if $n\geq N$, then $|f(x)-f_n(x)|<\epsilon$ for all $x\in K_1$. Writing $S=\sup_K |f|$, we in particular have $|f_n(x)|<|f(x)|+\epsilon\leq S+\epsilon$ for all $x\in K$, so $\sup_K |f_n(x)|\leq S+\epsilon$. I'll let you now think about how to choose $\epsilon$ to be able to reach the desired conclusion.