Arzela-Ascoli: Proof? Problem
Given a compact domain.
Regard the function space:
$$\mathcal{C}(\Omega):=\{f:\Omega\to\mathbb{C}:f\text{ continuous}\}$$
Consider a bounded family:
$$\mathcal{F}\subseteq\mathcal{C}(\Omega):\quad\|f\|_{f\in\mathcal{F}}<\infty$$
Then Arzela-Ascoli states:
$$\mathcal{F}\text{ precompact}\iff\mathcal{F}\text{ equicontinuous}$$
How to prove this from scratch?
Attempt
For a precompact family one finds:
$$\mathcal{F}\subseteq\mathcal{B}_\delta(g_1)\cup\ldots\cup\mathcal{B}_\delta(g_I)$$
So one can always pick one close enough:
$$f\in\mathcal{F}:\quad|f(x)-f(z)|\leq|f(x)-g_f(x)|+|g_f(x)-g_f(z)|+|g_f(z)-f(z)|<\varepsilon\quad(x\in B_\delta(z))$$
Conversely, prove for a sequence:
$$f_n\in\mathcal{F}:\quad\|f_{m'}-f_{n'}\|\to0$$
For a compact domain one finds:
$$\Omega\subseteq B_\delta(a_1)\cup\ldots\cup B_\delta(a_I)$$
Bolzano-Weierstrass gives a subsequence:
$$|f_n(a_i)|_{n\in\mathbb{N}}<\infty:\quad|f_{m'}(a_i)-f_{n'}(a_i)|\to0$$
Take as threshold:
$$m',n'\geq N':=\max_{i=1\ldots I}N'_i$$
So one can again always pick one close enough:
$$x\in\Omega:\quad|f_{m'}(x)-f_{n'}(x)|\leq|f_{m'}(x)-f_{m'}(a_x)|+|f_{m'}(a_x)-f_{n'}(a_x)|+|f_{n'}(a_x)-f_{n'}(x)|<\varepsilon$$
Is this proof correct or do I miss something?
Discussion
Moreover, why does the usual proof exploit separability before?
(For example see wiki: Arzela-Ascoli: Proof)
Sure for a proposition on its own:
$$\Omega\text{ separable}:\quad|f_n(x)-f(x)|\to0$$
$$\Omega\text{ precompact}:\quad\|f_n-f\|\to0$$
But why both together in a single proof?
 A: Consider any separable metric space $X$ and a complete metric space $Y$. Then endow $C(X,Y)$ with the compact convergence topology: a sequence $(f_n)$ converges to $f\in C(X,Y)$ if and only if for every compact subset $K$ of $X$, $f_n\mid K$ converges uniformly to $f\mid K$. You can show this is actually metrizable, so that compact and and sequentially compact are equivalent properties, and hence precompact and sequentially precompact are too. 
What Arzela-Ascoli says is that a family $\mathscr F$ of $C(X,Y)$ is precompact if and only if it is locally uniformly equicontinuous -- every point $x\in X$ has a neighborhood $V$ where  $\mathscr F\mid V$ is uniformly equicontinuous, and pointwise bounded, i.e. every set ${\rm ev}_x(\mathscr F)$ is bounded in $Y$ for each $x\in X$. 
It seems to me that when one puts it in this way, the proof is very natural and easy to remember. 
Consider a sequence $(f_n)$ in $\mathscr F$. First, take a dense countable subset $A=\{a_1,\ldots\}$. The general diagonal argument gives a subsequence $g_j=f_{i_j}$ such that $g_j(a_i)$ converges for each $a_i$ as $j\to\infty$. Now pick a compact set $K$, and take $\varepsilon >0$.
Since $\mathscr F$ is locally uniformly equicontinuous, each point $x$ in $K$ has a neighborhood $V_x$ where $\mathscr F \mid V_x$ is  is uniformly equicontinuous. Since $K$ is compact, we can find finitely many $V_1,\cdots ,V_s$ that cover $K$, and where $z,y\in V_x$ implies $d(f(z),f(y))<\varepsilon$ for any $z,y\in V_x$ and $f\in\mathscr F$. 
Since $A$ is dense, we can find $a'_i=a_{i_j}\in V_j$ for each $j$. I claim that $(g_j)$ converges uniformly in $K$. Indeed, pick any $z\in K$. If $z$ is in $V_i$ we have that 
$$d(g_n(z),g_m(z))\leqslant d(g_n(z),g_n(a_i))+d(g_n(a_i),g_m(a_i))+d(g_m(a_i),g_m(z))<3\varepsilon$$
Since $a_i,z\in V_i$, we have that $d(g_n(z),g_n(a_i))$ and $d(g_m(a_i),g_m(z))$ are both $<\varepsilon$ and since $g_j(a_i)$ converges we can pick $N$ so that $n,m>N$ gives $d(g_n(a_i),g_m(a_i))<\varepsilon$. This means that if $n,m>N$ and $z\in K$, $$d(g_n(z),g_m(z))<3\varepsilon$$
Which means $(g_n)$ converges uniformly in $K$. 
A: The separability argument does not lose any generality. The reason is that a compact metric space is totally bounded (which you are using), and a totally bounded metric space is separable. So if you assume that the domain is compact then you get that it is separable for free.
Here's a proof of the totally bounded implies separable argument:
Suppose $X$ is a totally bounded metric space, and consider covering $X$ by $N_n$ balls of radius $1/n$ which are centered at $\{ x_{i,n} \}_{i=1}^{N_n}$. Then the set $\{ x_{i,n} : i \in \{ 1,\dots,N_n \},n \in \mathbb{N} \}$ is a countable dense subset of $X$.
I find the separability argument convenient because it constructs the limit: we extract a subsequence which converges pointwise to a function $f$ on the dense subset, and then the limit function is the unique continuous extension of $f$ to the whole space. Equicontinuity then guarantees that this extension actually exists and must be the uniform limit of the original sequence.
