Prove the following statement:
Let $X$ be a set and $\mathcal{B} \subseteq \mathcal{P}(X)$. Then $\mathcal{A}:= \{\bigcap \mathcal{E}\mid \mathcal{E} \subseteq \mathcal{B}, |\mathcal{E}| \in \mathbb{N}\}$ is a basis for a unique topology on $X$.
I have already proved that the following two statements are equivalent:
(1) $\mathcal{A} \subseteq \mathcal{P}(X)$ is a basis for a unique topology on $X$
(2) $\forall A,B \in \mathcal{A}: \forall x \in A \cap B:\exists C \in \mathcal{A}: x \in C \subseteq A \cap B$ and $\mathcal{A}$ is a cover of $X$
So I'll use this in my attempt.
Take $\mathcal{E} := \emptyset$, the empty family. Then $X =\bigcap\mathcal{E} \in \mathcal{A}$. Hence, $X \subseteq \bigcup\mathcal{A} \subseteq X$ and it follows that $\mathcal{A}$ covers $X$.
Now, if $A := \bigcap \mathcal{E_1}, B := \bigcap \mathcal{E_2} \in \mathcal{A}$ with $\mathcal{E_i}\subseteq \mathcal{B} \quad(i=1,2)$, then
$$A\cap B = \bigcap \mathcal{E_1} \cup \mathcal{E_2}$$ and $\mathcal{E_1} \cup \mathcal{E_2} \subseteq \mathcal{B}$, a finite union, so that for every $x \in A\cap B$, we can simply take $C:= A \cap B$ and this completes the proof.
Is this correct?