Let $X$ be a compact metric space, and $(C(X),d_{\infty})$ the space of continuous functions. Let $D\subset{X}$ be a dense subset, and $\{{f_n\}}_{n \in N}$ a equicontinuous sequence of functions from $C(X)$. Show that if $\exists f\in C(x)$ such that $f_n(x)\rightarrow f(x)$ pointwise $\forall x\in D$, then $f_n(x) \rightarrow f(x)$ uniformly.
As $X$ is compact, $f$ is a uniform continous function. Then for $\varepsilon/3$ the exist a $\delta_1$ such that if $d(x,y)<\delta_1$ then $d(f(x),f(y))<\varepsilon/3$ for all $x,y \in X$.
For the same reason, and using the equicontinuity of $\{{f_n\}}_{n \in N}$, there exist a $\delta_2$ such that if $d(x,y)<\delta_2$ then $d(f_n(x),f_n(y))<\varepsilon/3$ for all $x,y \in X$ and for all $n$.
Taking $\delta=min\{{\delta_1,\delta_2}\}$ then, considering compactness $X$ can be written as a finite union of $B(x_i,\delta)$ with $i=1,2,...,m$ and for each $x_i$ we can choose a $q_i \in D$.
Then, using pointwise convergence for each $q_i$ we can find $n_i$ such that for all $n>n_i$ $d(f_n(q_i),f(q_i))<\varepsilon/3$
Finally taking $n_0=max\{n_1,n_2,...,n_m\}$ for all $n>n_0$, $d(f_n(x),f(x))<\varepsilon$ for all $x\in X$.
I am not sure if its right or clear :S. I know that pointwise convergence, equicontinuity and a compact set implies uniform convergence. But I am not sure how to deal whit point wise continuity over a dense subset. Thanks!