Why doesn't this argument show that every locally compact Hausdorff space is normal? In proving that a locally compact Hausdorff space $X$ is regular, we can consider the one-point compactification $X_\infty$ (this is not necessary, see the answer here, but bear with me).  Since $X$ is locally compact Hausdorff, $X_\infty$ is compact Hausdorff.  As a result, $X_\infty$ is normal.
Imitating the idea in my proof in the link above and taking into consideration the correction pointed out in the answer, let $A,B\subseteq X$ be two disjoint, closed sets in $X$, and consider $X_\infty$.  Since $X_\infty$ is normal, there are disjoint open sets $U,V \subseteq X_\infty$ such that $A\subseteq U$ and $B \subseteq V$.  Then considering $X$ as a subset of $X_\infty$, we can take the open sets $U \cap X$ and $V \cap X$ as disjoint, open subsets of $X$ that contain $A$ and $B$, respectively...or so I thought.  I see from the answers to this question that this does not succeed in proving that a locally compact Hausdorff space is normal, since this is not true.
So my question is simply: why does the above proof fail?
Thanks.

$A$ and $B$ are closed in $X$. They need not be closed in the compactification $X_\infty$. You could try to fix this by replacing them with their closures in $X_\infty$, but then these need not be disjoint.
