Proof. Let $\cb \subset 2^{X}$ be a base for $X$ and $A, B \subset X$ be disjoint closed sets. For each $x \in A$ and $y \in B$, there exists $U_{x}, V_{y} \in \cb$ such that $x \in \ol{U_x}\subset B^{c}$ and $y \in \ol{V_y}\subset B^{c}$. Since $\cb$ is countable, let $\seq{U_n}$ and $\seq{V_n}$ be enumerations of $\bracs{U_x|x \in A}$ and $\bracs{V_y|y \in B}$, respectively.

For each $n \in \natp$, let

then $U_{n}'$ and $V_{n}'$ are both open. Let

then $U \in \cn_{X}(A)$ and $V \in \cn_{X}(B)$. For each $m, n \in \natp$ with $m \le n$, $V_{n} \cap \bigcup_{j = 1}^{n} U_{j} = \emptyset$, so $U_{m} \cap V_{n} = \emptyset$. Likewise, if $m \ge n$, then $U_{m} \cap V_{n} = \emptyset$ as well. Therefore $U \cap V = \emptyset$.$\square$