Suppose $X$ is a topological space and $\{B_i\}$ form a base for the topology on $X$, where the $i$ run over some index set $J$.
$X$ is said to be compact if every open cover of $X$ contains a finite subcover of $X$.
Suppose you know that for every covering of $X$ by base elements, $X=\bigcup_{i\in I}B_i$, there exists a finite subcover $X=\bigcup_{i\in S}B_i$ where $S$ is a finite subset of $I$.
Does this then imply that $X$ is compact? If we have an arbitrary open covering of $X$, say $X=\bigcup_{j\in J} U_j$, then for each $j\in J$, there exists a covering of $U_j$ by some base elements $B_{j_i}$. Putting these together, we have $X=\bigcup_{i\in I,j\in J}B_{i_j}$, for which we know there is a finite subcover.
But does this then imply that our original cover of arbitrary open sets $U_i$ has a finite subcover?