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The finite-dimensional implication again follows from equivalence of norms and the Heine-Borel theorem. In an infinite-dimensional Banach space, inductively apply Riesz lemma to the proper closed finite-dimensional linear span $Y_n=\mathrm{span}(f_1,\dots ,f_n)$ to obtain a unit vector $f_{n+1}$ with $\mathrm{dist}(f_{n+1},Y_n)>1/2$. Therefore $\Vert f_n-f_m\Vert >1/2$ for distinct indices, precluding a Cauchy subsequence.

For completeness, Riesz lemma here follows by taking $z\notin Y$, setting $d=\mathrm{dist}(z,Y)>0$, choosing $y\in Y$ with $\Vert z-y\Vert <2d$, and putting $f=(z-y)/\Vert z-y\Vert$. Its distance from $Y$ exceeds $1/2$. This proves the closed unit ball is norm compact if and only if the Banach space is finite-dimensional. The printed hint's equality of all pairwise distances is unnecessary; the separation inequality is what the argument supplies.

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.

By the permitted sequential compactness of the product metric space $X\times X$, the sequence of pairs $({\varphi }^{(k)}x,{\varphi }^{(k)}y)$ has a convergent subsequence, and hence a Cauchy subsequence. Choose two of its indices $r<s$ sufficiently late thatThe distance-increasing assumption iterates: $d({\varphi }^{(r)}u,{\varphi }^{(r)}v)\ge d(u,v)$. With $n=s-r\ge 1$, use $u=x,v={\varphi }^{(n)}x$, and then $u=y,v={\varphi }^{(n)}y$, to obtainBoth summands are nonnegative, soThe use of paired iterates ensures that the same positive return time works for both points.