Satisfying the union of open sets axiom to prove unions of finite intersections of elements from a subbase form a topology

If I have a subbase $B$ of a topology $(X,T)$ (a subset of the power set of $X$ where the union of all elements in $B$ is $X$), then my topology notes state the collection of unions of finite intersections of elements of $B$ form a topology.

Here is my attempt at a proof:

1) By taking the union of every singleton of $B$ (considering a singleton as a finite intersection with itself), I get $X$, and by taking the empty union I get the empty set.

2) The union of a union of finite intersection of elements of $B$ is again a union of finite intersections of elements of $B$

3) The finite intersection of a union of finite intersection of elements of $B$ is again a finite intersection of elements for $B$ because...

Now for 3), I am a bit confused. Suppose I have two sets $A,C$ that are each unions of finite intersections of elements of $B$. How can I show that their intersection is again a union of finite intersection of elements of $B$ ? Hints and insights appreciated.

• I don’t believe it suffices to take the collection of finite intersections. I think you must take the collection of arbitrary unions of finite intersections. (However the collection of finite intersections form a base for a topology. Perhaps that was what was meant.) – spaceisdarkgreen Sep 13 '18 at 3:06
• @spaceisdarkgreen ok edited. I misread my notes. I am still confused however :). Hopefully this edit makes sense. – IntegrateThis Sep 13 '18 at 3:10
• What is your definition of a subbase of a topology? – Zest Sep 13 '18 at 3:17
• (a subset of the power set of X where the union of all elements in B is X) – IntegrateThis Sep 13 '18 at 3:19

For the third, use $$\bigcap_{i=1}^n \bigcup_{\alpha\in I_i}A^i_\alpha = \bigcup_{(\alpha_1,\ldots, \alpha_n)\in I_1\times\ldots\times I_n} \bigcap_{i=1}^n A_{\alpha_i}^i$$ (And check that I wrangled the sets right. Also remember $n=2$ suffices.)