Lemma 4.18.2. Let $X$ be a topological space, $\mathcal{U}\subset 2^{X}$ be locally finite, and $K \subset X$ compact, then $\bracs{U \in \mathcal{U}|U \cap K \ne \emptyset}$ is finite.
Proof. For each $x \in K$, there exists $N_{x} \in \cn(x)$ such that $\bracs{U \in \mathcal{U}|U \cap N_x \ne \emptyset}$ is finite. By compactness of $K$, there exists $X_{K} \subset X$ finite such that $K \subset \bigcup_{x \in X_K}N_{x}$. In which case,
\[\bracs{U \in \mathcal{U}|U \cap K \ne \emptyset}\subset \bigcup_{x \in X_K}\bracs{U \in \mathcal{U}|U \cap N_x \ne \emptyset}\]
$\square$