Proposition 4.16.4. Let $X$ be a Hausdorff space, $A, B \subset X$ be compact with $A \cap B = \emptyset$, then there exists $U \in \cn(A)$ and $V \in \cn(B)$ such that $U \cap V = \emptyset$.
In particular, if $X$ is compact and Hausdorff, then $X$ is normal.
Proof. For each $x \in A$ and $y \in B$, there exists $U_{x, y}\in \cn(x)$ and $V_{x, y}\in \cn(y)$ with $U_{x, y}\cap V_{x, y}= \emptyset$.
Fix $x \in A$. By compactness of $B$, there exists $B_{0}(x) \subset B$ finite such that $\bigcup_{y \in B_0(x)}V_{x, y}\in \cn(B)$. Let $U_{x} = \bigcap_{y \in B_0}U_{x, y}\in \cn(x)$ and $V_{x} = \bigcup_{y \in B_0(x)}V_{x, y}\in \cn(B)$, then $U_{x} \cap V_{x} = \emptyset$.
By compactness of $A$, there exists $A_{0} \subset A$ finite such that $\bigcup_{x \in A_0}U_{x} \in \cn(A)$. Let $U = \bigcup_{x \in A_0}U_{x} \in \cn(A)$ and $V = \bigcap_{x \in A_0}V_{x} \in \cn(B)$, then $A \cap B = \emptyset$.$\square$