Characterization for basis of generated topology 
Definition: Given a set $X$ and $\mathcal{B}$ a collection of subsets of $X$, the topology generated by $\mathcal{B}$ is the set:
$$[\mathcal{B}]=\bigcap_{\tau\in T}\tau$$
where $T=\{\tau \in \mathscr{P}(X):\tau$ is a topology on $X$ and $\mathcal{B}\subset\tau\}$
Question: Show that if $\mathcal{B}$ satifies:
  
  
*
  
*$\mathcal{B}$ covers $X$, that is, $\forall x \in X$, there is a $B\in \mathcal{B}$ such that $x\in \mathcal{B}$.
  
*$\forall B_1,B_2\in\mathcal{B}$ and $\forall x\in B_1\cap B_2$, there is a $C\in\mathcal{B}$ such that $x\in C\subset B_1\cap B_2$.
then $\mathcal{B}$ is a basis for the generated topology on $X$.

Almost everywhere I see the conditions in the question as the very definition of generated topology, and in the few places that said both definitions were equivalent, I didn't see a proof.
From 1, I got that (fairly obviously) that for every open set $A\in [\mathcal{B}$] we have:
$$A\subset\bigcup_{x\in A}B_x$$
where $x\in B_x\in\mathcal{B}$. But I got stuck when I tried to show the other inclusion for a subset $\mathcal{B}'\subset\mathcal{B}$.
I also tried by contradiction, only to- hit another wall.
Any tips are appreciated.
 A: The set $\tau$ of arbitrary unions of elements of $\mathcal{B}$ is a topology.
It is clear that $X\in\tau$, because of property 1; also $\emptyset\in\tau$ (union of the empty family).
It is also clear that $\tau$ is closed under finite intersections. Indeed, suppose
$$
U=\bigcup_{i\in I}B_i,
\qquad
V=\bigcup_{j\in J}C_j,
\qquad
(B_i,C_j\in\mathcal{B})
$$
Then
$$
U\cap V=\bigcup_{i,j}(B_i\cap C_j)
$$
so it's enough to prove that if $B,C\in\mathcal{B}$, then $B\cap C\in\tau$. This is obtained applying property 2.
Since $\tau$ is a topology and clearly it is included in every topology that includes $\mathcal{B}$, we are done.
A: Take an open set $A \in [\mathcal{B}]$ and take $x \in A$. Then, as $[\mathcal{B}]$ is the smallest topology that contains $\mathcal{B}$, it follows that in particular $A \in \{\cap \delta \colon \delta \subset \mathcal{B} \text{ and } \delta \text{ is finite}  \}$, because $\{\cap \delta \colon \delta \subset \mathcal{B} \text{ and } \delta \text{ is finite} \}$ form a topology that contains $\mathcal{B}$. 
By 2., it follows that for finite intersections, is still valid that there exists some $C \in \mathcal{B}$ such that $x \in C \subset A$.
So $\mathcal{B}$ is a basis for the topology $[\mathcal{B}]$
