A subspace of a compact regular space is locally compact

Question

Any hint for proving this? If $Y$ is a subspace of $X$, what I am able to find is a closed subset $V$ in $Y$, hence $\mbox{cl}_Y(V)$ is compact, whose closure is contained in a neighborhood of a point $x$, by regularity of $Y$. Restricted to $X$, this $V_x=V \cap X$ is closed. I dont see any way to prove that $V_x$ is compact in $X$ but this is the obvious path. Thanks a lot for any suggestion.

Answer 1

The result you wish to prove is not true. To give a counterexample I will use a characterization of locally compact subspaces of locally compact Hausdorff spaces found in this answer:

If a subspace $Y$ of a locally compact Hausdorff space $X$ is locally compact, then it is of the form $Y = F \cap U$ where $F \subseteq X$ is closed and $U \subseteq X$ is open.

Consider the compact regular space $X = [0,1]$, and the subspace $Y = \mathbb Q \cap X$. If $Y$ is locally compact, then there is a closed $F \subseteq X$ and an open $U \subseteq X$ such that $Y = F \cap U$. Since $Y \subseteq F$ and $Y$ is dense in $X$ and $F$ is closed, it follows that $F = X$, meaning that $Y = F \cap U = X \cap U = U$, which is impossible since $Y$ is not an open subset of $X$! Therefore $Y$ is not a locally compact subspace of $X$.

Answer 2

Let $X=[0,1]$ and $Y=X\cap \mathbb Q.$ Suppose $p\in U\subset Y$ where $U$ is a nbhd ,in $Y,$ of $p.$ Let $U\supset U'\cap Y$ where $U'$ is open in $X$ and $p\in U'.$ There exists $[a,b]\subset X$ with $a<b$ and $p\in [a,b]\cap Y$ and $[a,b]\subset U'.$ Now $$Cl_Y(U)=Y\cap Cl_X(U)\supset Y\cap Cl_X(U'\cap Y)\supset Y\cap Cl_X([a,b]\cap Y)=Y\cap [a,b].$$ Let $(x_n)_{n\in \mathbb N}$ be a strictly increasing sequence of irrationals and let $(y_n)_{n \in \mathbb N}$ be a strictly decreasing sequence of irrationals, with $a<x_1<y_1<b,$ and with $\lim_{n\to \infty}x_n=\lim_{n\to \infty}y_n\not \in \mathbb Q.$ Then $$\{[0,x_1)\cap Y, (y_1,1]\cap Y \}\;\cup \{(x_n,x_{n+1})\cap Y: n\in \mathbb N\}\;\cup \{(y_{n+1},y_n)\cap Y\}$$ is an open cover, in $Y,$ of $Cl_Y(U) ,$ without a finite sub-cover.

So no nbhd in $Y$ of any $p\in Y$ has compact closure in $Y.$