Prove $\overline{E}$ (closure) is compact

Question

In greater context, the question I'm trying to answer is:

Suppose $E \subset C([0,1])$ is equicontinuous and assume there is some $x_0 \in [0,1]$ and constant $M$ s.t. $|f(x_0)| \leq M, \ \forall f\in E$
Prove $\overline{E}$ (closure) is compact .

Just to clarify, does this mean that E is uniformly bounded at some $x_0$'s?

Also, I'm having trouble starting the proof,
I have a gut-feeling it will involve using Arzela-Ascoli's theorem, so I've been trying to show that $\overline{E}$ is bounded since I've successfuly shown that:

$E$ equicontinuous $\implies \overline{E}$ equicontinuous

and we know $\overline{E}$ is closed by def.

Would appreciate some hints regarding the problem

Answer 1

We can reformulate slightly the definition of (uniform, if you prefer the name) equicontinuity in the following way:

For each $\varepsilon > 0$ there exists $\delta > 0$ such that for any $f \in E$ and any $x_1, x_2$ such that $\lvert x_1 - x_2 \rvert \le \delta$ there holds $\lvert f(x_1) - f(x_2) \rvert \le \varepsilon$.

Specialize $\delta$ to correspond to $\varepsilon = 1$. Let $k \in \mathbb{N}$ be such that $\delta > \frac{1}{k}$. Then, for any $x \in [0, 1]$ one can get from $x_0$ to $x$ in at most $k$ steps of length $\delta$. Therefore

$\lvert f(x) \rvert \le M + k$, for any $f \in E$ and any $x \in [0,1]$.

Incidentally, the Arzelà–Ascoli theorem states that a (uniformly) bounded and (uniformly) equicontinuous subset of $C([0,1])$ has compact closure, so you do not need to check that $\overline{E}$ is equicontinuous.