A ${\bf subbasis}$ $\mathcal{S}$ for a topology on $X$ is a collection of subsets of $X$ whose union equals $X$. The ${\bf topology \; generated \; by \; the \; subbasis}$ $\mathcal{S}$ is defined to be the collection $\mathscr{T}$ of all unions of finite intersections of $\mathcal{S}$.
I want to check that $\mathscr{T}$ is indeed a topology. Following Munkres proof, he argues that it is enough to show that $\mathcal{B}$ (the collection of all finite intersection of elements $\mathcal{S}$ is a bases for $\mathcal{T}$) because then one can use Lemma 13.1(munkres) to prove that $\mathscr{T}$ , collection of all unions of elements in $\mathcal{B}$ is topology on $X$.
So, to check $\mathcal{B}$ is basis we need to verify two conditions:
$$ {\bf [1]} \;\;\; \forall x \in X, \exists B \in \mathcal{B} : x \in B $$
$$ {\bf [2]} \;\;\; if x \in B_1 \cap B_2 \implies \exists B_3\in \mathcal{B}: x \in B_3 \subset B_1 \cap B_2 $$
${\bf Proof.}$
${\bf [1]}$ Let $x \in X$ Since $X = \bigcup_{S \in \mathcal{S}} S$ so $x \in S$ for some $S$. Since $S = S \cap S$ then it belongs to $\mathcal{B}$. so we have found a basis element that contains $x$. $[1]$ is proved
${\bf [2.]}$ Consider now $B_1 = \bigcap^n S_i $ and $B_2 = \bigcap^m S_i'$ and say $x \in B_1 \cap B_2 $
Now, clearly $B_1 \cap B_2 $ is still ${\bf finite}$ intersections of elements of $\mathcal{S}$ so can we take $B_3 = B_1 \cap B_2$ in definition of bases?
Best Answer
Yes, the set of finite intersections of members of $\mathcal{S}$ is itself closed under intersections (of two or also finitely many) members so we can indeed take $B_3 = B_1 \cap B_2$ for any $x \in B_1 \cap B_2$.
It is in fact common to define $\cap \emptyset = X$ in this context (just as $\bigcup \emptyset = \emptyset$) so $X$ is also the intersection of a finite subcollection (the empty subcollection is finite) of $\mathcal{S}$. Then the only condition of $\mathcal{S}$ is also unnecessary (we only need it to have $X$ in the union of finite subcollections, as you saw).