Assume for contradiction that $X$ is infinite-dimensional. Starting with $M_0=\{0\}$, apply <Riesz lemma> recursively to the proper closed finite-dimensional subspaces
$$
M_{n-1}=\operatorname{span}\{x_1,\ldots,x_{n-1}\}
$$
to choose unit vectors $x_n$ satisfying
$$
\operatorname{dist}(x_n,M_{n-1})>\frac12.
$$
For $m<n$, the vector $x_m$ belongs to $M_{n-1}$, and therefore
$$
\lVert x_n-x_m\rVert>\frac12.
$$
Thus $(x_n)$ lies in the closed unit ball but has no <Cauchy subsequence>, hence no convergent subsequence. A compact metric space is <sequentially compact space>[sequentially compact], contradicting compactness of the unit ball. Therefore $X$ is finite-dimensional, proving that a <compact unit ball characterizes finite-dimensional normed spaces>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 22H / d / Solution

A normed vector space has compact closed unit ball if and only if it is finite-dimensional. In an infinite-dimensional space, repeated application of <Riesz lemma> produces unit vectors separated pairwise by a fixed positive distance, contradicting sequential compactness.

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