In a Banach space, the closed convex hull of a compact set is compact

Question

I'm trying to prove this result. Could you verify if my attempt is fine?

In a Banach space, the closed convex hull of a compact set is compact.

I post my proof separately as below answer. If other people post an answer, of course I will happily accept theirs. Otherwise, this allows me to subsequently remove this question from unanswered list.

Answer 1

Let $X$ is a Banach space and $K$ its compact subset. We want to prove that $\overline{\operatorname{conv} K}$ is compact. It suffices to show that $\overline{\operatorname{conv} K}$ is totally bounded.

Lemma 1: Let $(E, d)$ be a metric space. If $K \subset E$ is totally bounded, then so is its closure $\overline K$.

Proof: Fix $r>0$. There is $x_1, \ldots, x_n \in K$ such that $\{B(x_i, r/2)\}_{i=1}^n$ covers $K$. Clearly, $$ \overline K \subset \overline{\bigcup_{i=1}^n B(x_i, r/2)} \subset \overline{\bigcup_{i=1}^n \overline B(x_i, r/2)}= \bigcup_{i=1}^n \overline B(x_i, r/2) \subset \bigcup_{i=1}^n B(x_i, r). $$ The claim then follows.

By our Lemma 1, it suffices to show $\operatorname{conv} K$ is totally bounded. By Lemma 2 below, it suffices to prove that $K$ is totally bounded. This is trivially true because $K$ is compact.

Lemma 2: Let $(E, d)$ be a metric space. If $K \subset E$ is totally bounded, then so is its convex hull $\operatorname{conv} K$.