Every locally compact regular space contains a nonempty $G_\delta$-set In this page it is claimed that it is clear that every locally compact regular space contains a nonempty compact $G_\delta$-set. 
How is it clear?
 A: If by regular you mean what I would call regular and $T_1$, fix $x\in X$, and let $U$ be an open nbhd of $x$ with compact closure $K$. If $K=\{x\}$, then $K=U$ is open and hence a compact $G_\delta$. Otherwise, let $y\in U\setminus\{x\}$. $K$ is a compact Hausdorff space, so it is normal, and there is a continuous $f:K\to[0,1]$ such that $f(x)=0$ and $f(y)=1$. Let $C=f^{-1}[\{0\}]$; $C$ is a zero-set and hence a $G_\delta$, and since $C$ is a closed subset of $K$, $C$ is compact.
A: This answer is based on Brian M. Scott's answer. Since $X$ is compact and regular, by Wikipedia, it is completely regular. So $X$ can be equipped with a uniformity $\cal D$.
Assuming $X$ is nonempty,there is some $x\in X$. $x$ has a compact neighborhood $C$. Thus, $D=\overline{C}$ is a closed compact neighborhood of $x$. If $D=\overline{\{x\}}=\bigcap_{E\in \mathcal D}E[x]$, then $(\forall E\in \mathcal D)(D\subseteq E[x])$. But there is some open $E_0\in \mathcal D$ with $E_0[x]\subseteq D$. Thus, $D=E_0[x]$. Therefore, $D$ is open (and so $G_\delta$), compact and nonempty.
Otherwise, that is, if  $D\ne\overline{\{x\}}$, then there is some $y\in D \setminus\overline{\{x\}}$. Since $D$ is completely regular, there is a continuous function $f:D\to [0,1]$ with $f[\overline{\{x\}}]=\{0\}$ and $f(y)=1$. Let $E= f^{-1}[\{0\}]$. We have:
$$E=\bigcap_{n\in \Bbb N}f^{-1}\left(\left[0, \frac 1n\right)\right)$$
and so $E$  is a $G_\delta$-set. Also $E$ is closed and is contained in the compact set $D$. So $E$ is compact, contains $x$ and is a $G_\delta$-set.
