Proof. Let $U \subset X$ be open and $\seq{U_n}\subset 2^{X}$ be open and dense. Let $V_{0} = U$. For each $n \in \natp$, by density of $U_{n}$, there exists $x \in U_{n} \cap V_{n - 1}$. Since $X$ is regular (Definition 5.1.17/Proposition 4.16.4), there exists $V_{n}\in \cn^{o}(x)$ such that $x \in V_{n}\subset \ol U_{n}\subset U_{n} \cap V_{n-1}$. If $X$ is locally compact, choose $V_{n}$ to be precompact. If $X$ is completely metrisable, choose $V_{n}$ such that $\text{diam}(V_{n}) \le 1/n$.

Now, if $X$ is locally compact, then by the finite intersection property, $\bigcap_{n \in \natp}\ol{V_n}\ne \emptyset$. If $X$ is completely metrisable, then $\seq{V_n}$ is a Cauchy filter base, and converges to at least one point, so $\bigcap_{n \in \natp}\ol{V_n}\ne \emptyset$.

By Lemma 4.21.2, $X$ is a Baire space.$\square$