Let $E$ be a metric space and $F$ be a Banach space, $A\subset E$ dense. Let $(f_{n})$ be a squence of continuous and bounden functions from $E$ to $F$ such that the restriction of $f_{n}$ to $A$ converges uniformly. Show that $(f_{n})$ converges uniformly in $E$.
I think that using the Arzelà–Ascoli theorem I could solve this exercise, I believe that I have proved the relative compactness and the equicontinuity but I'm not sure if I'm on the right way.
Suppose $f_n \to f$ uniformly on $A$. Let $\epsilon >0$. There exists $n_0$ such that $\|f_n(x)-f(x)\| <\epsilon $ for all $x \in A$ if $n \geq n_0$. This gives $\|f_n(x)-f_m(x)\|<2\epsilon$ for all $x \in A$ if $n,m \geq n_0$. Since $A$ is dense in $E$ and $f_n$ is continuous for each $n$ we get $\|f_n(x)-f_m(x)\|\leq 2\epsilon$ for all $x \in E$ if $n,m \geq n_0$. (1) In particular, $(f_n(x))$ is a Cauchy sequence in $F$ for each $x \in E$. Let $f(x)=\lim f_n(x)$. This definition extends $f$ (orginally defined on $A$) to a function on $E$. Also, $\|f_n(x)-f(x)\|\leq 2\epsilon$ for all $x \in E$ if $n \geq n_0$. (This follows by letting $m \to \infty$ in (1)). We have proved that $f_n \to f$ uniformly on $E$.