$N$ complete metric space, if a sequence of continuous functions $f_n:M\to N$ conv. unif. in a subset $X$, then $f_n$ conv. unif. in $\overline {X}$ I need to prove the following:
Suppose $M$ a metric space and $N$ a complete metric space. Show that if a sequence of continuous functions $f_n:M\to N$ converges uniformly in a subset $X$, then $f_n$ converges uniformly in $\overline {X}$. (where $\overline{X}$ is the closure of $X$).
Well, first of all, what excatly does it means to say that a sequence of functions converges uniformly to $f$? I understand it by: for all $\epsilon>0$
there exists $n_0$ such that 
$$n>n_0\implies |f_n(x)-f(x)|<\epsilon$$
for all $x\in X$, and all $n>n_0$. 
Question: is this $n_0$ dependent of $\epsilon$ only?
Also, how do I relate this result to the closure of $X$?
I know of this theorem: suppose $f:A\to N$, if there exists $b=\lim_{x\to a} f(x)$, we have $b\in \overline{f(A)}$. Is this helpful?
 A: Yes, the $n_0$ only depends on $\varepsilon$, that is why it's called "uniform" (i.e. for all $x$ in $X$ at the same time!) convergence. If the $n_0$ is allowed to vary between $x$, it's called pointwise convergence.
You also omitted the statement "there exists $f: X \rightarrow N$ such that ..." and you probably know that a uniform limit of continuous functions is continuous, so $f$ is continuous on $X$, which might come in useful.
Now pick $p \in \overline{X}$. First you have to figure out what $f(p)$ should be. Indeed we can find sequences $(x_n)$ from $X$, that converge to $p$. And you would hope that $f(x_n) \rightarrow f(p)$: if the result would hold the supposed $f$ on $\overline{X}$ would also be continuous, if convergence were uniform.... But does the limit exist? Does it depend on the chosen sequence? 
Now is the time to use that $N$ is a complete metric space. Show that $f(x_n)$ is Cauchy in $N$ for every choice of sequence converging to $p$. Use unicity of limits in metric spaces to show that the limit does not actually depend on the specific sequence. Having done that, you have your candidate limit function on $\overline{X}$.  
