# Is every open subset of a locally compact Hausdorff space, locally compact in the subspace topology?

I tried proving it but I didn't use the fact of the subspace topology at all, therefore, I am doubting the correctness of the proof.

Let $X$ be a locally compact Hausdorff space and $Y$ an open subset of $X$. Since $X$ is a locally compact Hausdorff space, it is a regular space and since regularity is hereditary $Y$ is a regular space. Choose a random $x\in Y$ and a $U\subset Y$ such that $\overline{U}\subset Y$ with $x\in U$, we can do this because $Y$ is regular. since $\overline{U}$ is the closure of an open set in a locally compact space we know that $\overline{U}$ is compact in $X$. Therefore for every open cover of $\overline{U}$, there exists a finite subcover of $\overline{U}$. Because $Y$ is open in $X$, the intersection of (any open cover of $\overline{U}$ in $X$) and $Y$, would still be an open cover of $\overline{U}$ in $X$ therefore there exists a finite subcover for this open cover contained in $Y$. therefore $\overline{U}$ is compact in $Y$ which implies that $Y$ is locally compact.

$\mathbb R$ is locally compact. $U=\mathbb R$ is open in it. The closure $\overline {U}$ is not compact. So your assumtion that the closure of an open set in a locally compact space is compact is false.

Let O be an open subset and assume x in O.
Since X is regular,
exists open U with x in U, $\overline U \subseteq O.$
Since X is locally compact,
exists open V with compact closure and x in V subset U.
Hence the compact closure is a subset of O.
Thus O is locally compact.

Let $X$ be a locally compact Hausdorff space and $Y$ an open subset of $X$. Since $X$ is locally compact and Hausdorff $X$ is regular. Choose $x\in Y$, then there exists an open $U\subset X$ such that $\overline{U}$ is compact in $X$ with $x\in U$. Choose $V$ such that $\overline{V}\subset U$ and $\overline{V}\subset Y$ with $x\in V$ we can do this because $Y$ is regular. So $\overline{V}$ is a closed subset of a compact set, therefore $\overline{V}$ is compact in $X$. Therefore for every open cover of $\overline{V}$ there exists a finite subcover. The intersection of any open cover of $\overline{V}$ and $Y$ is open in $X$ because $Y$ is open, therefore for every open cover of $\overline{V}$ in $Y$ there exists a finite subcover in $Y$, therefore $Y$ is locally compact.

Let $x$ be in $Y$, and as $X$ is locally compact and regular and $Y$ is open, it has an open neighbourhood $V \subseteq Y$ with compact closure $\operatorname{cl}_X(V) \subseteq Y$. Then $V$ is open in $Y$, contains $x$ and $\operatorname{cl}_Y(V) = \operatorname{cl}_X(V) \cap Y = \operatorname{cl}_X(V)$ which is compact (in $X$ or $Y$ that does not matter, it's absolute). So $Y$ is locally compact Hausdorff.