Let $X$ be locally compact Hausdorff and $Y \subseteq X$ a closed subspace. The open subsets of $Y$ are precisely the open subsets of $X$ intersect $Y$. But given an open subset $U$ of $Y$, can we write $U = V \cap Y$ for some open subset $V$ of $X$? This seems plausible in "nice enough" spaces, but I can't think of a systematic way to build $V$ from $U$.
The problematic points are those in $U \cap \textrm{bd } Y$. For each $x \in U \cap \textrm{bd } Y$, I was thinking of adding some open neighborhood of $x$ to $U$. But to preserve $U = V \cap Y$, these open neighborhoods cannot intersect $Y \backslash U$. Since $X$ is regular, maybe I can take an open neighborhood of $x$ that does not intersect $\overline{Y \backslash U}$? I am not sure if this works as I am ignoring a lot of the fine details.
Edit: I realized my construction doesn't work in the case that $x \in \overline{Y \backslash U}$.
Consider $(X, \tau_X)$, an arbitrary topological space, and consider a subset $Y \subseteq X$.
We define the "subset topology" on $Y$ to be $\tau_Y = \{Y \cap V | V \in \tau_X\}$.
In other words, the open subsets of $Y$ are exactly those that can be written $Y \cap V$ for some open subset $V \subseteq X$.
So by definition, if we have any open subset $U \subseteq Y$, it is always possible to write $U = Y \cap V$ for some open $V$.