Exercise 34.4

The first implication it’s pretty easy to see. First, note that a compact hausdorff space $X$ can be considered as a subspace of it’s compactification of one point, thus it’s completely regular, since it’s completely regular then by the theorem of metrization of urysohn it has to be metrizable since it has a countable basis and it’s regular. The converse it’s not easy to see, in fact, to this point there is no counterexample of it in the book. We may think on the example of the subspace $(0,1)\times (0,1)$ of the ordered unit square. It is a hausdorff space and locally compact and has a metric induced by $d(a,b)=|a_2-b_2|$ if $a_1=b_1$
$d(a,b)=1$ if $a_1=b_1$ (Props to Camilo gallardo nuñez that came up with this counterexample).