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.