Alternate proof for Arzela-Ascoli

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Im trying to finish a beautiful excercise, which consist of giving an alternate proof for the following corollary of Arzela-Ascoli´s Theorem.

Given $X,Y$ metric spaces, $X$ compact, $Y$ complete, and let $\{f_n\}_{n\in \Bbb{N}} \subset C(X,Y)$, be an equicontinuous family of functions such that for every $x\in X$, $\{f_n(x)\}_{n \in \Bbb{N}}$ is relatively compact. Then by Arzela-Ascoli, there exist a subsequence of $(f_n)$ that converges uniformly.

So the alternate proof, is intended in the following way, take $\{x_m\}_{m\in \Bbb{N}}$ a dense subset of $X$. Show that there exist a subsequence of $(f_n)$, $(f_{n_{k1}})$ such that $(f_{n_{k1}}(x_1))$ converges. Show that there exist a subsequence of $(f_{n_{k1}})$, $(f_{n_{k1_{k2}}})$ such that $(f_{n_{k1_{k2}}}(x_2))$ converges.

Show that the diagonal sequence $f_{n_1},f_{n_{k1_{2}}},f_{n_{k1_{k2_3}}},...$ converges uniformly.

Ok, so I've done the following, since for every $x\in X$, $\{f_n(x)\}_n$ is relatively compact then, $\{f_n(x_1)\}_n$ is relatively compact and therefore there exist a subsequence $(f_{n_{k1}}(x_1))$ that converges in $Y$. Now, since $\{f_{n_{k1}}(x_2)\}_{k1}\subset \{f_{n}(x_2)\}_{n}$ then $\overline{\{f_{n_{k1}}(x_2)\}_{k1}}\subset \overline{\{f_{n}(x_2)\}_{n}}$ and therefore, $\overline{\{f_{n_{k1}}(x_2)\}_{k1}}$ is compact which implies that $\{f_{n_{k1}}(x_2)\}_{k1}$ is relatively compact and therefore, there exist a subsequence $(f_{n_{k1_{k2}}})$ such that $(f_{n_{k1_{k2}}}(x_2))$ converges.

Now, since $Y$ is complete, and $\{x_m\}_{m \in \Bbb{N}}$ dense, then if $f_n \to f$ uniformly in $\{x_m\}_{m \in \Bbb{N}}$ then it converges uniformly in $\overline{\{x_m\}_{m \in \Bbb{N}}}=X$.

Ok so I have to hyphotesis left to use, $X$ is compact and the family $\{f_n\}$ is equicontinuous, to show that the diagonal sequence, $f_{n_1},f_{n_{k1_{2}}},f_{n_{k1_{k2_3}}},...$ converges uniformly.

My intuition says that the $f$ to which the diagonal sequence converges, should be defined as $f(x_1)=\lim f_{n_{k1}}(x_1)$, $f(x_2)=\lim f_{n_{k1_{k2}}}(x_2),...$

However im not being able to use the compactness and equicontinuity (or uniform equicontinuity which is implied by the previous) to conclude the proof. How can I relate equicontinuity with convergence?

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Read the proof of Lemma 9.4.12 in H. H. Sohrab Basic Real Analysis 2nd edition (2014), pp. 448-449.

Reading the proofs of Lemmas 9.4.10 and 9.4.11 can be useful too.

For a discussion, I prefer to wait for your comments.