I am reading a textbook proving the Tychnoff's theorem, which states that any product of compact spaces is compact.
The proof goes like this:
Let $X=\prod_{\alpha \in A} X_\alpha$, where each $X_\alpha$ is compact. Let $\mathcal{D}$ be a family of subbasic sets of the form $\pi_\alpha^{-1}(U_\alpha)$, where $\alpha \in A$ and $U_\alpha$ is an open subset of $X_\alpha$. Suppose that no finite subfamily of $\mathcal{D}$ covers $X$. In view of the Alexander Subbase Theorem, it suffices to show that $\mathcal{D}$ does not cover $X$.
I wonder why it suffices to show that $\mathcal{D}$ does not cover $X$. The Alexander Subbase Theorem says that if every open cover of $X$ by sets in subbase has a finite subcover, then $X$ is compact. By the above proof, we can show that if $\mathcal{D}$ covers $X$, then there is a finite subcover. How can we show that for every open cover (not just $\mathcal{D}$ itself), there is a finite subcover?
If the space $X$ is not compact, there is an open cover, consisting of sets in a subbase, such that no finite subfamily covers $X$.
If you show that any family, with the property that no finite subfamily covers $X$, is not a cover, you have shown it is false that $X$ is not compact.