Close to a proof that every such equicontinuous sequence of functions converges uniformly. Let $X$ be a compact metric space; let $Y$ be a normed space; let $\{ f_{n} \} \subset \mathscr{C}(X, Y)$ be equicontinuous; and let $f_{n} \to g$ pointwisely for some $g: X \to Y$. Claim: We have $f_{n} \to g$ uniformly.
I think I am close to a proof of the claim. 
Let $\varepsilon > 0$. Since $\{f_{n}\}$ is equicontinuous by assumption, for every $x \in X$, by definition there is some $X$-open $U \ni x$ such that 
$|f_{n}(x) - f_{n}(y)| < \varepsilon$ for all $y \in U$ and all $n \geq 1$.
Since $X$ is compact by assumption, so by definition there are some $X$-open $U_{1}, \dots, U_{k}$ covering $X$; let $x_{i} \in U_{i}$ for all $1 \leq i \leq k$.
Since $f_{n} \to g$ pointwisely by assumption, by definition there is some $N \geq 1$ such that
$|g(x_{i}) - f_{n}(x_{i})| < \varepsilon$ for all $n \geq N$ and all $1 \leq i \leq k$.
I made use of all the assumptions, but, what is a thing I overlooked so that I cannot continue the above?

Continuing with what you did, let $y \in X$ be arbitrary. Then $y \in
U_i$ for some $i$. Hence,
\begin{eqnarray*}
&&|f_n (y) -f_m (y)| \\&\leq &|f_n (y)-f_n (x_i)|+|f_n
(x_i)-g(x_i)|+|g(x_i)-f_m(x_i)|+|f_m (x_i)-f_m(y)| \\ &<&4\epsilon
\end{eqnarray*}
for sufficiently large $n,m$ (where "sufficiently large" does not
depend on $y\in X$.
We have thus shown that the sequence $(f_n)_n$ is uniformly Cauchy. By
classical results, it thus has a (uniform) limit $h$ in
$C(X,\overline{Y})$, where $\overline{Y}$ is the completion of $Y$.
Since you know that the pointwise limit is given by $g$, we get $g=h$
and thus the desired uniform convergence.
I am sure that one can eliminate the whole "completion of $Y$
argument", but if you are satisfied with the argument as it is, I will
not do so :)
