Equivalence for sequentially compact space for Hausdorff spaces.

Question

Let $X$ be a Hausdorff space. Show that $X$ is sequentially compact if and only if for every closed set $A$ of $X$ and every continuous function $f:A \to \mathbb{R}$, $f$ has a maximum. I think this should follow from Urysohn's lemma but I don't really see how to approach this problem. I know that since $X$ is Hausdorff, we also get that $X$ is normal. So in particular, $X$ is a compact Hausdorff space, however I don't know if this would get me anywhere...

Answer 1

This is false: $\{0,1\}^{[0,1]}$ is compact Hausdorff (so every real-valued continuous function assumes a max) but not sequentially compact. $\beta \Bbb N$ (the Čech–Stone compactification of $\Bbb N$ is another such example.

In first countable Hausdorff spaces the equivalence (pseudocompactness and sequential compactness) does hold.