 Solution

Assume for contradiction that $X$ is infinite-dimensional. Starting with $M_0=\{0\}$, apply Riesz lemma recursively to the proper closed finite-dimensional subspacesto choose unit vectors $x_n$ satisfyingFor $m<n$, the vector $x_m$ belongs to $M_{n-1}$, and thereforeThus $(x_n)$ lies in the closed unit ball but has no Cauchy subsequence, hence no convergent subsequence. A compact metric space is sequentially compact, contradicting compactness of the unit ball. Therefore $X$ is finite-dimensional, proving that a compact unit ball characterizes finite-dimensional normed spaces.