Are one-point sets always compact in any topological spaces? From the definition of compactness, I think one-point sets are always compact in any topological space. But, I am not sure about my judgement. Am I correct?
 A: Since I was asked to repost my comment as an answer:
Any open cover of any topological space is a subset of the power set of the underlying set, and power sets of finite sets are finite. So all open covers of finite spaces are already finite; in other words, finite spaces only have finitely many open sets.
A: Every finite set is compact. This is because, one can always find a finite subcover, which can be proven inductively:
Let $X:=\{x_1, \dots x_n\}$ be a  finite set. Suppose that $\{U_x\}$ covers $X$. Take any $U_{x_1}$ that covers $x_1$. Consider $\{U_x\}\setminus U_{x_1}$. Then pick one that  covers $x_2$ if $U_{x_1}$ does not, etc.
A: Yes, one point set is always compact in any topological space, because it will be contained in an open set of any cover and that is the finite one.
A: Yes.
Suppose the open sets $U_i$ with $i$ in some index set $I$ cover your one-point set $\{x\}$.
Covering this set means that $x\in U_i$ for some $i\in I$.
Therefore your one-point set is contained in a single open set of your open cover.
This is certainly a finite subcover!
Notice that the argument did not use the fact that the sets $U_i$ are open.
This is a symptom of generality:
It is true for any sets $U_i$ and therefore for any topology of the ambient space at all.
