If f is uniformly continuous in a metric space, there is limit in set closure $ f:X \subset \mathbb{R}^m \to \mathbb{R}^n  $ uniformly continuous in $ X $, show that $ \forall a \in \overline{X} $ exists $ \lim_{x \to a} f(x) $. 
Is enough to say:
Since f is uniformly continuous then is also uniformly continuous in all coordinate functions. With that since $ f_1, \ldots, f_n $ is uniformly continuous we have $ | f_i(x) - f_i(a) | < \epsilon_i, \; 1 \le i \le n $. Using the limit definition we have $ \lim f_i(x) = L_i \equiv | f_i(x) - L_i | < \epsilon_i $. If we take $ f_i(a) = L_i $. We won.
Is that ok?

It is not ok because we don't know if $f({\bf a})$ exists. For example, we could be taking $f:(0,1)\to \Bbb R$ and ${\bf a} = 1$. 
Since ${\bf a} \in \overline{X}$, we have a sequence ${\bf x}_n$ in $X$ such that ${\bf x}_n \to {\bf a}$. Then I claim that $\lim_{{\bf x}\to {\bf a}}f({\bf x}) = \lim_{n \to +\infty} f({\bf x}_n)$ (can you check this?). Meaning, I have to prove that $f({\bf x}_n)$ does converges to something. But I can't say it is to $f({\bf a})$. The point is, if $({\bf x}_n)_{n \in \Bbb Z_{>0}}$ converges, then it is a Cauchy sequence. And uniformly continuous functions take Cauchy sequences to Cauchy sequences (do you know how to prove it?). So $(f({\bf x}_n))_{n \in \Bbb Z_{>0}}$ is Cauchy, and since $\Bbb R^n$ is complete, it converges. 
