Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 1 / 10F / a / Solution

This is always true: it is the Heine-Cantor theorem. If uniform continuity failed, there would be an $\epsilon >0$ and sequences $x_n,y_n\in X$ such thatBy sequential compactness of a compact metric space, some subsequence $x_{n_k}$ converges to $x\in X$. The triangle inequality gives $y_{n_k}\to x$. Continuity then makes both image subsequences converge to $f(x)$, contradicting their separation by $\epsilon$.