Newbie in topology. Feels like I'm doing mental gymnastics and missing the point. I have the following two definitions and a lemma from munkres
Which are helpful for problem 2(a) of Section 26 below
The following solution makes sense to me and was my first instinct, however, I wanted to start off differently somehow to ensure it would have an open covering.
"Any open set in a collection that covers a set covers all but finitely many points of the set. Therefore, a finite subcollection covers it all"
When we are showing that a subspace is compact, is it only the implication that we have a finite subcovering which matters? How can I be assured that this arbitrary subspace of the the real numbers under the finite complement topology will actually have an open covering? Of course the whole space would "cover" this subspace, but the definition of a covering seems to require equality not containment. Can I even use the whole space as an assured "covering"? Any help much appreciated!
Yes, you only need to show that for any open cover, there exists a finite subcover – not that a particular open cover exists.
This is not really a concern of compactness. As you noted there is the trivial covering using the whole space, but in general having fewer open sets makes compactness an easier property to verify. For example, in the cofinite topology from the Munkres exercise, the open sets are fairly simple to describe, so it is easy to classify the compact sets. Compare this to $\mathbb{R}^n$, where classifying the compact sets is much more involved.
This is somewhat of a technicality but Lemma 26.1 essentially just means that verifying compactness of $Y$ with the subspace topology from $X$ is the same as verifying compactness of $Y\subseteq X$ with the topology of $X$. By the definition of topology there is always at least one open cover: the covering by the whole space, so you can be assured that the collection of coverings is nonempty.