Uniform limit of holomorphic functions

For a function $f: D \to \mathbb{C}$ to be holomorphic, by Morera's theorem it's enough that for every triangle $\Delta \subset D$, we have $\int_{\partial \Delta} f = 0$. Now, if $f_n \to f$ uniformly, it's easy to show that $\int_{\partial \Delta} f_n \to \int_{\partial \Delta} f$, but as $f_n$ are holomorphic, $\int_{\partial \Delta} f_n = 0$, so the limit is of course also 0.

Note that it's actually enough to assume almost uniform convergence, i.e. uniform convergence on all compact subsets $K \subset D$.

You only need that $f_n\to f$ uniformly on every compact subsets of $D$. It's a well-known fact that $f$ is then continuous. The idea is to use Morera's theorem.

Let $\Delta\subset D$ be a closed triangle. Since each $f_n$ is holomorphic, by Cauchy's theorem, you have $\displaystyle\int_{\partial\Delta} f_n(z)dz=0$ for all $n$.

$\partial\Delta$ is a compact subset of $D$, so you know that $f_n\to f$ uniformly on $\partial\Delta$.

So you get, for all $n$, $$\left|\int_{\partial\Delta} f(z)dz\right|=\left|\int_{\partial\Delta} (f(z)-f_n(z))dz\right|\leq\mathrm{length}({\partial\Delta})\sup_{z\in\partial\Delta}|f(z)-f_n(z)|$$

By letting $n\rightarrow\infty$, you find that $\displaystyle\int_{\partial\Delta} f(z)dz=0$.

By Morera's theorem, $f$ is holomorphic.

You've already seen an approach using Morera's theorem from the other excellent answers. For a slightly more concrete demonstration of why $f$ is complex differentiable, you can use the fact that every $f_n$ satisfies Cauchy's integral formula, so by uniform convergence $f$ also satisfies Cauchy's integral formula. This allows you to differentiate $f$ by differentiating the integral.