Let $X$ be a topological space, and let $\mathcal V$ be a basis.
If I want to prove that $X$ is compact then it is clear to me that it is enough to prove that every open cover of elements of $\mathcal V$ has a finite subcover.
My question is:
Is this also the case if $\mathcal V$ is not a basis but is a subbasis?
I really don't see that.
Further I am motivated to ask this by a proof that I met here (lemma 3.1 page 6).
It shows that the set of prime filters of a bounded distributive lattice equipped with the topology generated by the collection $\{\phi(a)\mid a\in L\}\cup\{\phi(b)^{\complement}\mid b\in L\}$ is compact. Here $L$ denotes the lattice and $\phi(a)$ is the set of prime filters that contain $a$ as element (see definition 1.3). Further it is evident that the set $\{\phi(a)\cap\phi(b)^{\complement}\mid a,b\in L\}$ serves as basis of the topology.
The proof starts by saying that it is enough to prove it for covers of "basic sets" (so far I agree), but then only gives a proof for a cover of sets that come from the subbasis. This puzzles me.
If the answer to my posed question is "no" then another question:
Is the proof that I quoted okay?
Thank you in advance for your time.