You misunderstood something. Since $\tau$ only includes subsets of $X$. It does not include $\mathcal P(X)$ itself as an element.
Instead it includes every element of $\mathcal P(X)$ (read: subset of $X$) which can be written as the union of these singletons. Which subsets are these?
Yes, this is correct. Every subset of $X$ is open in this case.
As for the final question, yes, it is possible to have bases of different cardinalities, for example by taking $\{\{a\},\{b\},\{c\},\{a,b\}\}$ you obtain another basis for the same topology, and of course the topology itself is always a basis for itself.
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.
Best Answer
In your condition (2), it is important to point that we must have $x \in B_3$, such that $x \in B_3 \subset B_1 \cap B_2$.
Let's organize our thoughts.
You have fixed a topological space $(X, {\cal T})$. A collection ${\cal B}\subset {\cal T}$ is a basis if and only if satisfies conditions (1) and (2). End.
Now take a set $X$, without topology, and consider the power set $\wp(X)$. If you take ${\cal B}\subset \wp(X)$ and ${\cal B}$ verifies conditions (1) and (2), then there exists a unique topology on $X$ which has $\cal B$ as a basis. End.
Ok?