The forward direction is trivially proven. For the converse, suppose towards contradiction that is not compact but every subbasic open cover from has a finite subcover. Let denote the set of all open covers of lacking a finite subcover partially ordered by inclusion. By Zorn's lemma, has a maximal element , which itself is an open cover of admitting no finite subcover, By maximality, if is an open set such that then has a finite subcover for some finite .

Now clearly does not cover , for if it did it would have a finite subcover. Thus there exists some not covered by . But since does cover , there exists some with . Since is a subbasis, there must exist some finite collection of subbasic open sets such that . Now , for otherwise would cover . Defining as above and , it follows is a finite subcover of for any . Let and . Now for any , covers iff , whence for each and thus , implying is a finite subcover of , contradicting . Therefore must be compact.