Proof of the Arzelà–Ascoli Theorem

I'm stuck on a particular line of the proof of The Arzelà–Ascoli Theorem.

In lectures, we have:

$1.$ Defined equicontinuous as:

Let $X$ be a metric space, $C(X) = \{f: X \rightarrow \mathbb{R}\text{ continuous} \}$ the space of continuous functions, $S \subset C(X)$. Let $x \in X$ be a point. Then $S$ is equicontinuous at $x$ if $\forall \varepsilon > 0$, $\exists \delta > 0$ such that $y \in B(x, \delta)$, $f \in S$ $\implies$ $|f(x) - f(y)| < \varepsilon$.

And obviously $S$ is equicontinuous if it is equicontinuous at all points of $X$.

$2.$ Stated The Arzelà–Ascoli Theorem as:

Suppose that $X$ is a compact metric space and $S \subset C(X)$ is a subspace.

Then, $S$ is compact $\iff$ $S$ is closed, bounded, and equicontinuous.

In the proof of the forward direction:

Closed and bounded are clear, so it remains to show that $S$ is equicontinuous. We already know that $S$ is totally bounded, so let $\varepsilon > 0$ and fix $x \in X$. Then $\exists F \subset S$ finite such that $S \subset \bigcup_{f \in F}B(f, \frac{\varepsilon}{3})$.

Since $F$ is equicontinuous...

This is the line that I get stuck at, why is $F$ equicontinuous?

• Because it is finite. A finite subset of $C(X)$ is equicontinuous (in general this is not true, but since $X$ is compact, for all $f \in C(X)$, $f$ is uniformly continuous). – Crostul Mar 30 '16 at 10:16

Recall that for all $f \in C(X)$, $f$ is uniformly continuous because $X$ is compact.
Given $F \subset C(X)$ a finite set, we show that it is equicontinuous. Fix $\varepsilon > 0$. We need to find som $\delta > 0$ such that some condition is satisfied.
For all $f \in F$, by uniform continuity of $f$, there exist $\delta_f$ such that etcetera. Your required $\delta$ is $\delta = \min_{f \in F} \delta_f$.