Are these two definitions of basis equivalent? Lecture note definition

Let $(X, \mathcal{T})$ be topological space, A $basis$ of $\mathcal{T}$ is a collection $\mathcal{B}$ of open sets satisfying the following:
For each open set $U$ and for each element $x \in U$, there exists a set $\beta$ such that $x \in \beta$ and $\beta \subset U$

Most textbook's definition:

If $X$ is a set, a basis for a topology on $X$ is a collection $\mathcal{B}$ of subsets of $X$ such that

*

*for each $x \in X$. there is at least one basis element $B$ containing $x$


*if $x$ belongs to intersection of two basis element $B_1$ and $B_2$, then there is a basis element $B_3$ such that $B_3$ containing $x$ and is contained in the intersection of $B_1$ and $B_2$

If so, how do I prove it ?
 A: Keep in mind that the first definition refers to a basis for a given topology, and the second one to a set $\mathcal{B}$ that will work as a basis for a topology, not given, but determined or generated by $\mathcal{B}$.
The fundamental thing about a basis, say $\mathcal{B}$, is that any open set in the topology for which $\mathcal{B}$ is a basis is a union of elements of $\mathcal{B}$.
Notice in the first definition, the fact that for any open set $V$ you have a basic set $U_{x}$ sandwiched between every point $x$ and $V$ makes every open set a union of elements of the basis, i.e: $V=\bigcup_{x\in V} U_{x}$, a union of basic sets.
The second definition is usually accompanied by something a long the lines of:

The topology $\mathcal{T}$ generated by the basis $\mathcal{B}$ is defined by: A set $V$ is called open in $X$ (that is $V\in\mathcal{T}$) if for each $x\in V$ there is a $B_{x}\in\mathcal{B}$ such that $x\in B_{x}\subset V$.

Notice that in this case too any open set in the topology $\mathcal{T}$ generated by the basis $\mathcal{B}$ will be an union of elements of $\mathcal{B}$. The fact that a topology defined in the manner described above is actually well defined (I mean actually closed under unions and finite intersections) depends purely on that the two requisite for being a basis are met. I see it like this:
We want to create a topology where certain sets of our selection are open, to make it a topology we have to close our set under arbitrary unions and finite intersections. But if we're lucky and our set meets the two requirements given in your definition for a basis, closing under arbitrary unions is enough, closing under finite intersections is already taken care off. This is seen through an argument similar to the one presented above, given $B_{1},B_{2}\in\mathcal{B}$, for any $x\in B_{1}\cap B_{2}$ there is a $B_{x}\in\mathcal{B}$ such that $x\in B_{x}\subset B_{1}\cap B_{2}$, effectively making this set ($B_{1}\cap B_{2}$) an union of elements of $\mathcal{B}$. Recapulating: I want to include $B_{1}\cap B_{2}$ in my topology, but if $\mathcal{B}$ is a basis, I've already included it when I added all the unions of all elements of $\mathcal{B}$ to my topology.
Just trying to give you some intuition on the point of talking about bases, maybe you can try and formalize a proof for the equivalency you want now.
