Lemma 4.18.3. Let $X$ be a topological space, $\mathcal{U}\subset 2^{X}$ be locally finite, then $\bracsn{\ol{U}|U \in \mathcal{U}}$ is also locally finite.
Proof. For each $x \in X$, there exists $N_{x} \in \cn^{o}(x)$ such that $\bracs{U \in \mathcal{U}|N_x \cap U \ne \emptyset}$ is finite. Since $N_{x}$ is open, for any $U \in \mathcal{U}$, $N_{x} \cap U = \emptyset$ implies that $N_{x}^{c} \supset \ol{U}$. Thus $\bracsn{U \in \mathcal{U}|N_x \cap \ol U \ne \emptyset}$ is finite as well.$\square$