Prove any class of subsets of non-empty set is a subbase for a unique topology of on the set. Prove the statement, any class $\mathcal{A}$ of subsets of a non-empty set $X$ is the subbase for a unique topology $\tau$ on $X$, that is, finite intersections of members of $\mathcal{A}$ form a base for a topology $\tau$  on $X$.
Proof.
Define $\mathcal{B}$ as the class of finite intersections of members of $\mathcal{A}$, then we must show that $$X= \cup \{ B : B \in \mathcal{A} \}$$ and that for any $G,H \in \mathcal{B} $, $G \cap H$ is the union of members of $\mathcal{B}$.
Note that $X \in \mathcal{B} $ since by definition $X$ is the empty intersection of members of $\mathcal{A}$ so $$X=\cup \{ B : B \in \mathcal{B} \}$$
Also, if $G,H \in \mathcal{B}$ then $G$ and $H$ are finite intersections of members of $\mathcal{A}$, therefore $G \cap H$ is also finite intersection of members of $\mathcal{A}$ and so belongs to $\mathcal{B}$.
$\mathcal{B} $ is a base for some topology $\tau$ on $X$ for which $\mathcal{A}$ is the subbase. $\tau $ is also unique by a known theorem.
I don't understand a part of this proof, specifically, why is it true that $X \in \mathcal{B}$.
Consider this case, if $X:=\{a,b,c,d \}$ and $\mathcal{A}:=\{  \{a\} , \{ a,b \}  \}$ then since $\mathcal{B}$ is a class of finite intersections of members of $\mathcal{A}$ then $\mathcal{B} =\{ \{a \}, \{a,b \} \}$, but $X \notin \mathcal{B}$.
What is my mistake? Can you explain why it's the case that $X \in \mathcal{B}$?
 A: Let's take a look at the definition of $\mathcal B$:

$\mathcal B$ is the class of finite intersections of members of $\mathcal A$.

Here's perhaps a better way of saying it:

$\mathcal B$ is the class of subsets of $X$ which can be written as the intersection of the members of some finite subset of $\mathcal A$.

So for any a finite subset $\mathcal F \subset \mathcal A$ you get an element $B \in \mathcal B$ by taking the intersection.
$$B = \{x \in X \mid \forall F \in \mathcal F , x \in F\}
$$
Applying this to $\mathcal F = \emptyset$, we get
$$B = \{x \in X \mid \forall F \in \emptyset , x \in F\} 
$$
Since $\emptyset$ has no elements, the condition in this set builder notation is vacuously true, and so
$$B = \{x \in X \} = X
$$
That's what they mean by saying that "$X$ is the empty intersection of members of $\mathcal A$".

I'll add that while this is formally valid, nonetheless many authors go out of their way to avoid this "vacuous truth" argument, by including the explicit requirement $X \in \mathcal B$ as part of the definition of $\mathcal B$ (as is suggested in the other answer).
A: Your first mistake is that for $\mathcal A$ to be a subbase of a topology $\mathcal T$ you must verify that

The collection of open sets consisting of all finite intersections of
elements of $\mathcal A$, together with the set $X$, form a basis for T.

So $X$ should be added to the elements of $\mathcal A$.
Your second mistake is that the class $\mathcal B$ of finite intersections of members of $\mathcal{A}$ includes all members of $\mathcal A$ as for any element $A \in \mathcal A$ you have $A = A \cap A$.
