It is known that not every locally compact Haussdorff space is normal, see for example

But it seems that the following is true, I just want to make sure I am not making any mistake:

**Lemma** Let $X$ be a locally compact Haussdorff space and let $K, W \subset X$ be compact with $K \cap W = \emptyset$. Then, there exists open sets $K \subset U, W \subset V$ such that $U \cap V =\emptyset$.

*Proof:*

We have $K \subset (X \backslash W)$. Since $K$ is compact, $X \backslash W$ is open and $X$ is locally compact Haussdorff space, by Theorem 2.7 in Rudin, "Real and Complex Analysis" we can find an open set $U$ (with compact closure) such that $K \subset U \subset \bar{U} \subset X \backslash W$.

Let $V:= X \backslash \bar{U}$. Then $V$ is open and satisfies the above conditions.

\qed

Is this correct? I am a bit uneasy, especially since the proof only uses the fact that $W$ is closed, compactness is only needed for $K$.