I have learned that the Arzela-Ascoli theorem essentially requires the separability of the domain in order to use the diagonal argument. While I read the paper "The Kolmogorov-Riesz Compactness Theorem" written by Harald Hanche-olsen and Helge Holden, I saw that Theorem 2 (The Arzela-Ascoli theorem) does not require the separability of the domain. In fact the proof in this paper uses total boundedness instead of sequential compactness, however, I claimed that the proof through the sequential compactness can be also completed without the trick of Cantor's diagonal. Here is my attempt, and I could not find errors. Am I right?
Statement: Let $X$ be a compact topological space. Let $Y$ be a complete metric space. Let $C(X,Y)$ be a space of continuous functions with the topology of uniform convergence. Let $F\subset C(X,Y)$.
Suppose $F$ is pointwise equicontinuous, i.e. for every $\varepsilon$, there is an open neighborhood $U_x$ of $x$ such that $y\in U_x$ implies $d(f(x),f(y))<\varepsilon$ for every $f\in F$. Suppose $F$ is pointwise relatively sequentially compact, i.e. for every $x\in X$, every sequence in the set $\{f(x):f\in F\}$ has a Cauchy subsequence.
Then, $F$ is uniformly relatively sequentially compact, i.e. every sequence in $F$ has a uniformly Cauchy subsequence.
Proof) Suppose $\{f_n\}_{n\in\mathbb{N}}$ is a sequence in $F$. We want to show this sequence has a uniformly Cauchy sequence.
Let $\varepsilon>0$.
Step1: Make an open cover of $X$. Let $V_x$ be an open neighborhood of $x$ such that $y\in V_x$ implies $d(f(x),f(y))<\varepsilon/3$ for every $f\in F$. Then, $\{V_x\}_{x\in X}$ is an open cover of $X$.
Step2: Take a finite subcover. Since $X$ is compact, we can take a finite subcover $\{V_{x_k}\}_k$ of X.
Step3: Take a subsequence. We can take a subsequence $\{g_n\}_n$ of $\{f_n\}_n$ such that $\{g_n(x_k)\}_n$ is Cuachy for all $k$. (It does not use diagonal argument because the index $k$ is finite!!) Define $N_k$ as a natural number such that $n,m>N_k$ implies $d(g_n(x_k),g_m(x_k))<\varepsilon/3$.
Step4: Show $\{g_n\}_n$ is uniformly Cauchy. For the $\varepsilon$, there is a natural number $N:=\max_k N_k$ such that $n,m>N$ implies $d(g_n(x),g_m(x))<\varepsilon$ for every $x\in X$, because there is $k$ with $x\in V_k$ such that $d(g_n(x),g_m(x))\le d(g_n(x),g_n(x_k))+d(g_n(x_k),g_m(x_k))+d(g_m(x_k),g_m(x))\le\varepsilon.$
(The first and third term is due to the equicontinuity(Step1) and the second term is due to the Cauchyness(Step3))
For extra question, I think the proof of Theorem 2 in the paper can be extended for noncompact domain. In this case, we should set the topology of compact convergence on the continuous function space. Is it possible?