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).

Let $C$ be the positively oriented unit circle.  By the Cauchy integral formula, for $z$  in the open disk
$$ f_n(z) = \dfrac{1}{2\pi i} \oint_C \dfrac{f_n(\zeta)\; d\zeta}{\zeta-z}$$
so that taking the limit as $n \to \infty$,
$$  f(z) = \dfrac{1}{2\pi i} \oint_C \dfrac{f(\zeta)\; d\zeta}{\zeta-z}$$
Now the difference quotient
$$ \eqalign{\dfrac{f(z) - f(w)}{z-w} &= \dfrac{1}{2\pi i} \oint_C \frac{f(\zeta)}{z-w} \left( \frac{1}{\zeta - z} - \frac{1}{\zeta - w}\right) \; d\zeta\cr
& = \dfrac{1}{2\pi i} \oint_C \dfrac{f(\zeta) \; d\zeta}{(\zeta-z)(\zeta-w)}\cr
& \to \dfrac{1}{2\pi i} \oint_C \dfrac{f(\zeta) \; d\zeta}{(\zeta-w)^2} \ \text{as}\ z \to w}$$
which says $f$ is complex differentiable at $w$ with derivative $$f'(w) = \dfrac{1}{2\pi i} \oint_C \dfrac{f(\zeta) \; d\zeta}{(\zeta-w)^2}$$ 
