TITLE: Isn't every subset of a compact space compact?
QUESTION [5 upvotes]: Let $\mathfrak{C}$ be an open cover of a topological compact space $X$, and let $\mathfrak{B}\subseteq \mathfrak{C}$ be its finite subcover. Then every subset of $X$ also has the same cover and subcover! Shouldn't every such subset also be compact then?
Motivation: I read somewhere that every closed subset of a compact space is compact. I wonder why it shouldn't be the case for every subset of $X$, rather than just the closed ones.

REPLY [5 votes]: The big issues here are

That open sets of a subspace are not always open in the larger space.
That the open sets you add to get a cover of the whole space may have to be very large and contain infinitely many of the subspaces open sets. These large open sets are often preferentially chosen in the finite subcover, like in the one point compactification of a discrete space.