Is a basis for a given topology always closed under finite intersections?

Clearly, the system of all open intervals of lengths $1/n$, $n=1,2,\dots$, is a base for the usual topology on $\mathbb R$. $$S=\{(a,a+1/n); a\in\mathbb R\}.$$ This base $S$ is not closed under finite intersections: $$(0,1/2)\cap(x,x+1/2)=(x,1/2)$$ for every $x\in(0,1/2)$. If we choose $x$ which is not of the form $1/2-1/n$, then this intersection does not belongs to $S$.

Your first comment is false. Every open set will be a union of basis elements, but not necessarily finitely many basis elements.

The usual topology on $\mathbb{R}$, for example, is generated by the basis $S$ consisting of all open intervals. But there are many open sets that are not a union of finitely many intervals. For example there are unbounded open sets.

The answer to the question in the title is also no. For example, choose a single element of $S$ above and remove it to obtain $S'$. $S'$ still generates the same topology (since the interval you threw away can be easily obtained as a union of two smaller intervals) but it is missing an intersection in $S'$ (since the interval you threw away can be easily obtained as an intersection of two larger intervals).