Lemma 5

Let $X$ be a regular space such that, for each open covering $\mathscr{R}$ of $X$, there exists a (not necessarily open) locally finite covering $\mathscr{U}$ of $X$ which refines $\mathscr{R}$. Then for each open covering $\mathscr{R}$ of $X$ there exists a locally finite closed covering $\mathfrak{F}$ of $X$ which refines $\mathscr{R}$.

Let $\mathfrak{R}$ be any open covering of $X$. For each $x \in X$ there is an open set $U \in \mathfrak{R}$ which contains $x$, and therefore (since $X$ is regular) an open neighbourhood $V_x$ of $x$ such that $\overline{V}_x \subset U$. The family $\mathscr{B}$ formed by the $V_x$ is an open covering of $X$, hence by hypothesis there is a locally finite covering $\mathscr{B}'$ which is finer than $\mathscr{B}$. Let $\mathscr{F}$ be the family of closures of the sets of $\mathscr{B}$. Since the covering $\mathscr{B}'$ formed by the $\overline{V}_x$ is finer than $\mathfrak{R}$, and since $\mathscr{F}$ is finer than $\mathscr{B}'$, it follows that $\mathscr{F}$ is a closed covering of $X$ which refines $\mathfrak{R}$. But also $\mathscr{F}$ is locally finite, because if an open set does not meet a set $B \in \mathscr{B}$, then it does not meet its closure $\overline{B}$ either.