= Solution
Suppose the continuous $f$ were not <uniformly continuous>. There would be an $\varepsilon_0>0$ and pairs $x_n,y_n\in X$ with $d(x_n,y_n)<1/n$ but $|f(x_n)-f(y_n)|\geq\varepsilon_0$. <sequentially compact space>[Sequential compactness] gives a subsequence $x_{n_j}\to x\in X$. The <triangle inequality> gives $y_{n_j}\to x$ as well. <Continuity> at $x$ then gives $f(x_{n_j})-f(y_{n_j})\to0$, a contradiction. Thus \b[$f$ is uniformly continuous].
It need not be <Lipschitz continuous>. On the sequentially compact interval $[0,1]$, the continuous function $f(x)=\sqrt x$ has
$$
\frac{|f(x)-f(0)|}{|x-0|}=\frac1{\sqrt x}\to\infty\quad(x\downarrow0).
$$
No finite Lipschitz constant exists.
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