Is a dense, locally compact subspace always open in a compact Hausdorff space?

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Here is this week's POTW:

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Let $X$ be a compact Hausdorff space. If $X$ contains a dense, locally compact subspace $S$, show that $S$ is open in $X$.
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This week's problem was solved correctly by Janssens. You can read his solution below.

Spoiler

Let $p \in S$ be arbitrary. By local compactness there exists an open subset $U$ of $S$ such that $p \in U$ and the closure $\overline{U}^S$ of $U$ in $S$ is compact in $S$, hence in $X$ as well. Since $X$ is Hausdorff, it follows that $\overline{U}^S$ is closed in $X$, so
$$
\overline{U} \subseteq \overline{U}^S,
$$
where the left-hand side (without the superscript) denotes the closure in $X$.

Secondly, since $U$ is open in the subspace topology of $S$, there exists an open set $O \subseteq X$ such that $U = O \cap S$. We show that $O \subseteq S$. We note that
$$
O \subseteq \overline{O \cap S}.
$$
(Indeed, let $x \in O$ be arbitrary and let $B \subseteq X$ be any open set with $x \in B$. Since $O$ is open and $S$ is dense in $X$, the open set $B \cap O$ intersects $S$. Hence $B \cap (O \cap S) \neq \emptyset$ so $x \in \overline{O \cap S}$.) Next, using the two previously displayed equations in order, we obtain the inclusions
$$
O \subseteq \overline{O \cap S} = \overline{U} \subseteq \overline{U}^S \subseteq S.
$$
This shows that $p$ is an interior point of $S$ with respect to $X$, so $S$ is open in $X$.

Remarks: Compactness of $X$ is not needed. Also, I recall this problem from a lecture I enjoyed in the past, so I do not want to pretend the above solution is entirely my own.