I have the following theorem:
Let $f_n: X \to \Bbb{R}^k$ be a sequence of functions defined in a compact space $X$. If the collection $\{f_n\}_{n \in \Bbb{N}}$ is equicontinuous and if $\forall x \in X$ the sequence $\{ f_n(x)\}$ is a bounded sequence in $\Bbb{R}^k$ then there is a subsequence of $\{f_n\}$ that converges uniformly to a continuous function $f$.
This Theorem is presented without proof and my attempt is presented below. I would like to be sure my proof is correct and I would also like to be given a pointer as to how to finish.
Let $H = \{f_n\}_{n \in \mathbb{N}} \subset C(X, \mathbb{R}^k)$. By hypothesis $H$ is equicontinuous and pointwise totally bounded.
$X$ is compact hence by a previous theorem I get for free that $H$ is totally bounded. Because $H$ is a metric space and $\mathbb{R}^k$ is a complete metric space, I know that $\overline H$ is complete. With this and the fact that $H$ is totally bounded I will try to recursively build a Cauchy subsequence $\{f_{n_k}\} \subset H$ and then show it converges uniformly to some function $f$ on $\overline H$ because $\overline H$ is complete.
Cover $H$ with open sets of diameter less than $1$. It is immediate that at least one of said open sets contains an infinite number of terms of our sequence $\{f_n\}$. Let it be $A_1$ and pick $f_{n_1} \in A_1$. If there isn't said $A_1$, then $\{f_n\}$ admits a constant subsequence and the result follows immediately.
Now cover $A_1$ with open sets of diameter less than $1/2$. Again, there exists some $A_2$ with an infinite number of terms of our sequence. Pick $f_{n_2} \in A_2$ with $n_2 > n_1$.
Keep going like this to build a sequence $\{f_{n_k}\}$ that is trivially a Cauchy sequence. Now all there is left to show is that it converges uniformly to some $f$, which I don't see how to do. If it is the case that this sequence converges uniformly, then said $f$ is continuous by another theorem I have.