 Solution

We prove uniform continuity by contradiction. If it fails, there is ${\epsilon }_0>0$ such that, for every $n\ge 1$, we can choose $x_n,y_n\in X$ withBy sequential compactness, pass to a subsequence with $x_{n_k}\to x\in X$. The triangle inequality givesBy continuity, both $f(x_{n_k})$ and $f(y_{n_k})$ converge to $f(x)$. A second triangle inequality therefore givescontradicting the lower bound. Thus the requested $\delta$, independent of $x,y$, exists for every $\epsilon >0$.