TITLE: Any space $X$ with the indiscrete topology is compact.
QUESTION [5 upvotes]: Let $\tau_X = \{ \varnothing, X \} $. Let $A \subseteq X$, Let $O$ be an open cover of $A$. Since topology on $X$ is finite, then $O$ must finite too. Obviously, any subcover $O'$ of $O$ must be finite, therefore $X$ must be compact by definition.
Is this correct? It seems obvious to write it down or probably im missing somthing? any feedback? thanks

REPLY [7 votes]: You don't even need to consider $A\subseteq X$. What you must show is that any open cover of $X$ contains at least one finite subcover of $X$. But note that the only open cover of $X$ is the singleton $\{X\}$, and it trivially contains a finite subcover: $\{X\}$ itself.
Addition; pointed out by @RickyDemer: $\{\varnothing,X\}$ is another open cover of $X$, as $X\subseteq \varnothing\cup X=X$, but it is a finite cover as well.