TITLE: Compact space, locally finite subcover
QUESTION [10 upvotes]: This is an exercise from a topology book.
$X$ is a topological space. If every open cover of $X$ has a locally finite subcover, then $X$ is compact.
What I have tried: I have considered the cover composed of all open subsets of $X$; this has a locally finite subcover. So we can conclude that in this space every point is in a finite number of open sets.

REPLY [4 votes]: I think it is worth converting Mike F's answer into a direct proof.
Given a nonempty topological space $X$ with the property described in the question and any open cover $\mathscr{U}$, we will construct a finite subcover.  Pick a point $p\in X$, and pick some $U\in\mathscr{U}$ containing $p$.  Then $\mathscr{U}':=\{U\cup V:V\in\mathscr{U}\}$ is an open cover of $X$ with a point-finite subcover $\mathscr{V}'$.  Since every element of $\mathscr{U}'$ contains $p$, $\mathscr{V}'$ is finite, so the set of $V$s that give rise to $\mathscr{V}'$ together with $U$ itself forms a finite subcover of $X$.