Let $X$ be a topological space. Show that if there exists an embedding $i:X\rightarrow Y$ where $Y$ is compact Hausdorff and $i(X)$ is open in $Y$, then $X$ is locally compact and Hausdorff.
I know that it would be equivalent to show that there exists a one-point compactification of $X$, which I assume is easier. Also, the fact that $i$ exists implies that a compactification exists, so $j:X\rightarrow Z$, where $Z$ is compact Hausdorff and $\overline{j(X)}=Z$. I'm not sure how to use the fact that $i(X)$ is open in $Y$ to show that there is a one-point compactification. Can I assume that $Z=X^+$ is a one-point compactification?
"$X$ locally compact, Hausdorff $\Rightarrow$ there exists a one-point compactification" is definitely easy. But it's not clear to me that "$X$ embeds into $Y$ with $Y$ compact $\Rightarrow$ $X$ has a one-point compactification" is all that easy.
But it's quite easy to show that $X$ is locally compact and Hausdorff using the given hypothesis, so I would suggest you do that.
First $Y$ is Hausdorff. So for any two points in $X$ we have neighborhoods in $Y$ separating them. The intersection of those neighborhoods with $i(X)$ gives neighborhoods in $X$.
Second, since $Y$ is compact and Hausdorff, it is locally compact. And a subspace of a locally compact Hausdorff space is locally compact if it is the difference of closed sets. Since we have $i(X) = Y\setminus (Y\setminus i(X)),$ therefore $i(X)\cong X$ is locally compact.