Open subspaces of locally compact Hausdorff spaces are locally compact Let $X$ be locally compact and Hausdorff. I want to prove the following:

If $Y\subset X$ is open then $Y$ is locally compact.

I have proved that closed subsets of $X$ are locally compact, but how can one prove this?
I also want to use these two lemmas to conclude the following:

If $Y\subset X$ is locally closed then $Y$ is locally compact.

 A: I know this proof doesn't really use the preliminary results you wanted to use, but I can't resist.
Let $X$ be locally compact Hausdorff. Then $X$ has a one-point compactification $\alpha X=X\cup\{\infty\}$ and a Stone-Cech compactification $\beta X$.  The identity map on $X$ extends to a map $\beta X\to \alpha X$, and the remainder $\beta X \setminus X$ maps to ${\infty}$.  Thus $X$ is open in $\beta X$.  If $Y$ is open in $X$ then $Y$ is also open in $\beta X$. It is easy to see that open subsets of compact Hausdorff spaces are locally compact, so there you go!
A: Let $X$ be a locally compact Hausdorff space.
To show that an open subspace $Y$ of $X$ is itself locally compact, given $x \in Y$ fix open neighbourhoods $U$ and $V$ of $x$ in $X$ such that


*

*$\overline{U}$ is compact. (It is possible to do this because $X$ is locally compact.)

*$\overline{V} \subseteq Y$. (It is possible to do this because locally compact Hausdorff spaces are regular.)


Now $U \cap V$ is an open neighbourhood of $x$ in $X$ (and also in $Y$). Furthermore


*

*as $\overline{U \cap V} \subseteq \overline{V} \subseteq Y$, then $\operatorname{cl}_Y ( U \cap V ) = \overline{U \cap V} \cap Y = \overline{U \cap V}$.

*as $\overline{U \cap V}$ is a closed subset of the compact $\overline{U}$, it is also compact.


Since you already know that closed subspaces of $X$ are locally compact, it now easily follows that locally closed subspaces of $X$ are locally compact. Recall that being locally closed in $X$ is equivalent to being of the form $U \cap F$ where $U \subseteq X$ is open and $F \subseteq X$ is closed. So $F$ is locally compact as a closed subspace of $X$, and $U \cap F$ is locally compact as an open subspace of $F$.

It may be worth mentioning that locally compact subspaces of Hausdorff spaces are locally closed. (See this question and its answer.) And so in a locally compact Hausdorff space, being a locally compact subspace and being a locally closed subspace are equivalent notions.
A: The previous proofs were too complicated and rather confusticate the problem.
Instead you can simply follow your nose:

We follow the definition of a locally compact space as it should be:*
Every point has a neighborhood basis of compact subsets.
Having settled our grounds, let's get straight to the proof:
Consider a locally compact space and an open subspace $X\subseteq Y$.
Let $x\in U\subseteq X$ be a neighborhood of $x\in X$.
Assume wlog that $U$ is open in $X$ (e.g. $U=X\cap V$).
So also $U$ open in $Y$ as by assumption $X\subseteq Y$ open!
By assumption $Y$ is locally compact.
So we may sandwich this in $x\in L\subseteq U$ with $L$ a compact nbhd of $x\in Y$.
Since this took place in $L\subseteq X$, the restricted neighborhood remains $L\cap X=L$.
This defines the desired compact nbhd.

*Two small remarks for common agreement:

*

*Generally these are not open and possibly not closed.

*Hausdorffness does not play any role in our proof. It only serves to characterise (for the original space $Y$) that the existence of single compact neighborhood may be sufficient for local compactness.

