Uniform limit of continuous analytic functions is analytic 
Let $f_n:\mathbb C \to \mathbb C$ be functions that are continuous on the closed unit disk and analytic on the open unit disk. Suppose that $f_n$ converges uniformly to some $f$ on the the closed unit disk. Prove that $f$ is analytic on the open unit disk.

I'm looking for a way to prove this without resorting to complex analysis theorems.
Here are my thoughts so far : pick $z_0$ in the open unit disk and $\epsilon >0$. There's some $N$ such that $n\geq N\implies ||f_n-f||<\epsilon$. For each $n\geq N$, there's some $\delta_n>0$ and $(a_{n,k})_k$ such that
$\forall z\in B(z_0,\delta_n), f_n(z) = \sum_{k=0}^\infty a_{n,k}(z-z_0)^k$
Since $||f_n-f_m||$ is Cauchy, I'm wondering if it's possible to build a convergent subsequence of $(a_{n,k})_n$ for each $k$. If that is true, I expect the limits to be the coefficients of $f$ as a power series.
Another issue is that it could be that $\delta_n \to 0$.
I haven't used yet that the $f_n$ are continuous on the boundary of the circle.
Any help is appreciated (preferably without complex analysis).

So far, there is no difference in your approach and what one might try for functions analytic on $[0,1].$ But by the Weierstrass approximation theorem, the result fails there.
