Every metrizable space $(X, \tau)$ is perfectly normal: for every [closed set](/page/Closed%20Set) $C \subset X$, there exists a [continuous](/page/Continuity) function $f: X \to [0, 1]$ with $C = f^{-1}(\{0\})$. In particular, [metrizable spaces](/page/Metrizable%20Space) are normal and hence satisfy Urysohn's Lemma.