This is a famous result and I would like to give it a try based on what I have learned from Brezis's Functional Analysis.
Let $(X, |\cdot|)$ be a normed space and $B:= \{x \in X \mid |x|\le1\}$ the closed unit ball. Then $B$ is compact if and only if $X$ is finite-dimensional.
Could you have a check on my attempt? I'm happy to receive your comments/suggestion.
The proof relies on the following facts about weak topology $\sigma(X, X^*)$.
Lemma: Let $S := \{x\in X \mid |x|=1\}$ be the unit sphere. If $X$ is infinite-dimensional, then $\overline S^{\sigma(X, X^*)} = B$. [The proof of can be found at page 59]
A direct consequence of the Lemma is that the unit sphere is never closed in $\sigma(X, X^*)$ if $X$ is infinite-dimensional.
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If $X$ is finite-dimensional, then $X$ is topologically isomorphic to $\mathbb R^d$ for some $d \in \mathbb N^*$. A subset of $\mathbb R^d$ is compact if and only if it is closed and bounded. It follows that $B$ is compact.
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Let $B$ be compact in norm topology. Because $S \subset B$ is closed in norm topology, $S$ is compact in norm topology. It follows that $S$ is compact in $\sigma(X, X^*)$ which is Hausdorff. It follows that $S$ is closed $\sigma(X, X^*)$. It follows from our Lemma that $X$ is finite-dimensional.
Best Answer
@DanielWainfleet has provided another approach via Riesz lemma. Below is my formulation of his idea.
Any finite-dimensional subspace of $E$ is a closed proper subspace of $E$. Fix $\alpha \in (0, 1)$.
First, we pick an arbitrary $x_0 \in B$. By Riesz lemma, there is $x_1 \in B$ such that $d(x_1, \operatorname{span} \{x_0\}) \ge \alpha$. Inductively, we pick $x_{n+1} \in B$ such that $$ d(x_{n+1}, \operatorname{span} \{x_0, \ldots, x_n\}) \ge \alpha. $$
It follows that $(x_n) \subset B$ is not Cauchy and thus not convergent. Hence $B$ is not compact.