The kind of convergence you refer to is (compact) normal convergence.
This can be shown using Morera's theorem in conjunction with Cauchy's integral theorem, which states:
A continuous function $f$ is analytic in an open domain $\Omega$ iff for every
closed, zero-homological, rectifiable curve $\gamma$ the line integral
$\oint_\gamma f(z) d z = 0$. (A zero-homological loop is a loop with
winding number zero.)
We can use this result to prove the desired property:
Continuity of the limit follows immediately from the uniform convergence. It remains to verify the key element of Cauchy's integral theorem, namely that all line integrals $\oint_\gamma f(z) dz = 0$ for the limit.
Provided that $f_n\to f$ uniformly on every compact set, you have in particular $f_n\to f$ uniformly on every closed loop $\gamma$ (of finite length). Hence, for any zero-homological loop $\gamma$, and every $\epsilon>0$ there exists some $N(\epsilon)\in \mathbb N$ such that $|f_n(z)-f(z)|<\epsilon$ for all $n>N(\epsilon)$. If $l(\gamma)$ denotes the length of $\gamma$, we then obtain
$$\left|\oint_\gamma f_n(z) dz - \oint_\gamma f(z) dz\right| \le l(\gamma)\epsilon,\quad\forall n>N(\epsilon).$$
Since for every $n$, the integral $\oint_\gamma f_n(z) d z = 0$, this proves that $\oint_\gamma f(z) d z = 0$. Hence $f$ is analytic.