 Solution

First prove the suggested density fact. If $Z$ is dense in $Y$, $V$ is open and $y\in V$, then every open neighborhood $W$ of $y$ has $W\cap V$ nonempty open. It meets $Z$, so $W$ meets $V\cap Z$. HenceIdentify $X$ with its dense embedded image in the hausdorff compactification. For $x\in X$, local compactness supplies an open neighborhood $U$ in $X$ and a compact set $K\subseteq X$ with $x\in U\subseteq K$. Choose an open set $V$ in $Y$ with $V\cap X=U$. The inclusion into $Y$ is continuous, so $K$ is compact in $Y$ and therefore closed, since $Y$ is Hausdorff. The density fact now yieldsThus every point of the embedded $X$ has a $Y$-open neighborhood contained in $X$. Consequently