Spaces with finite bases are compact

By the definition every open subset of $X$ can be written as a union of some subcollection from $B$. Now $B$ is finite so there is finitely many subcollections of $B$ and therefore $X$ has finitely many open subsets. In other words the topology $\tau$ on $X$ is a finite set.

So if we take any collection $S$ of open subsets (it doesn't even have to be a cover, any collection) then it already is finite! Simply because $S$ is a subset of $\tau$ which is finite. So just put $S'=S$ and you're done.

Side note: you typically don't assume that members of the (sub)cover are disjoint. This is not required for compactness. And indeed in most cases (e.g. $X$ connected) this can't even happen.

For each point $x$ in $X$, pick an open neighbourhood $U_x\in S$ of $x$. By definition of basis, for each $x$ and corresponding $U_x$, there is an open basis neigbourhood $V_x\subseteq U_x$ of $x$ (i.e. $V_x\in B$).

There are only finitely many elements in the basis. Therefore, only finitely many distinct basis elements could have been chosen. Pick points $x_1,x_2,\ldots,x_n \in X$ such that $V_{x_1}, V_{x_2}, \ldots, V_{x_n}$ are all the chosen basis elements. Now $S' = \{U_{x_1}, U_{x_2}, \ldots,U_{x_n}\}$ is a finite subset of $S$.

Take an arbitrary $x\in X$. Then there must be an $x_i$ such that $V_x = V_{x_i}$. Since $x\in V_x = V_{x_i} \subseteq U_{x_i}$, $x$ is covered by $S'$. This shows that $S'$ covers all of $X$, and we are done.