Why proving this is sufficient? I was reading here in page 3 the proof that the finite intersection of elements in the sub-basis generate a basis, but I don't understand why is enough to say that the intersection of elements in $B$ (the basis) is an element of $B$, 
Can someone clarify please?  
Thanks a lot in advance.
 A: There are two conditions for $\mathcal{B}$ to be basis: every point in your space should be contained in a basic open set, and any point contained in two basic open sets $B_1,B_2 \in \mathcal{B}$ should be contained in a basic open set contained in $B_1 \cap B_2$. For the second condition to hold it is sufficient (but in general not necessary) to show that $B_1 \cap B_2$ is a basic open set. This is what they are showing.
A: One of the conditions that a family of sets must meet in order to be a basis, is that for any two basis elements, say $U$ and $V$, and any $x\in U\cap V$, there is a basis element $W$ such that $x\in W$ and $W\subseteq U\cap V$.  If the intersection of any two basis elements is itself a basis element, this condition is trivially satisfied.
A: In order for a collection $\mathcal{B}$ of subsets to be a basis, two axioms must be satisfied.  The first is that the union of elements of $\mathcal{B}$ must be the whole space.  The other is that for any two $B_1,B_2\in \mathcal{B}$ and any $x\in B_1\cap B_2$, there exists a $B_3\in \mathcal{B}$ such that $x\in B_3$.  If $B_1\cap B_2\in \mathcal{B}$ already, then this second axiom is trivially satisfied.  
