= Solution
{c}
We prove <uniform continuity> by contradiction. If it fails, there is $\varepsilon_0>0$ such that, for every $n\geq1$, we can choose $x_n,y_n\in X$ with
$$
d(x_n,y_n)<1/n,\qquad \rho(f(x_n),f(y_n))\geq\varepsilon_0.
$$
By <sequential compactness>, pass to a <subsequence> with $x_{n_k}\to x\in X$. The <triangle inequality> gives
$$
d(y_{n_k},x)\leq d(y_{n_k},x_{n_k})+d(x_{n_k},x)\longrightarrow0.
$$
By <continuity>, both $f(x_{n_k})$ and $f(y_{n_k})$ converge to $f(x)$. A second <triangle inequality> therefore gives
$$
\rho(f(x_{n_k}),f(y_{n_k}))\leq\rho(f(x_{n_k}),f(x))+\rho(f(x),f(y_{n_k}))\longrightarrow0,
$$
contradicting the lower bound. Thus the requested $\delta$, independent of $x,y$, exists for every $\varepsilon>0$.
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