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


[here](https://mathoverflow.net/questions/53300/locally-compact-hausdorff-space-that-is-not-normal)


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 $K$.



Separating compact sets in locally compact spaces
Yes, this is known and you don't need local compactness even as long as $X$ is completely regular (locally compact spaces are completely regular). 


I think that the best way to prove is to go via the Čech–Stone functor, which is available precisely for completely regular spaces. Consider $\beta X$ and note that $K, W\subset X\subset \beta X$ remain compact in $\beta X$. 


