If $X$ admits a Hausdorff compactification, then $X$ is locally compact? I read that if $X$ admits a Hausdorff compactification, that is a compact Hausdorff $C$ such that $X$ is homeomorphic to an open dense subset of $C$, then $X$ is locally compact. Why is that?
 A: Because if $C$ is compact Hausdorff, any open subset of it is locally compact (and Hausdorff). So if $X$ embeds as an open (dense) subset of $C$ it is also locally compact and Hausdorff as its embedded image is. 
A: The statement you wrote is false, but it becomes true if you instead consider the one point compactification, rather than an arbitrary one.
Let $ X' = X \cup \{\infty\} $ denote the one point compactification of $ X $. Let $ x \in X $. We suppose $ X' $ is Hausdorff, so let $ U $ and $ V $ be disjoint open sets such that $ x \in U $ and $ \infty \in V $. Then $ V $ is an open neighborhood of infinity, so by defintion of the topology of $ X' $, there is some compact closed set $ K $ in $ X $ such that $ V = X'-K $. Then as $ U \cap V = \emptyset $, $ U \subseteq X' - V = K $. Then as $ K $ is closed in $ X $, $ \overline{U} \subseteq K $. Then as $ K $ is compact and $ \overline{U} $ is closed, $ \overline{U} $ is compact, and is therefore a compact neighborhood of $ x $. Thus, $ X $ is locally compact.
In fact, the converse is true as well. A locally compact Hausdorff space admits a Hausdorff one point compactification.
