TITLE: Weakening paracompactness condition
QUESTION [5 upvotes]: Let $X$ be a topological space such that every open cover has a finite refinement. Then is $X$ compact, or is there a counterexample?
Let $X$ be a topological space such that every open cover has a locally finite subcover. Then is $X$ compact, or is there a counterexample?

REPLY [3 votes]: For 1., clearly yes: if a cover has a finite refinement, then the sets it is a refinement of also cover $X$, and form a finite subcover.
For 2., see Nuno's comment on the other answer.