Prove the proposition.

Question

this is my first year studying topology and I was given a proposition but not its proof so I was wondering how i would make it but i didn't reach anything. It states the following:

Let $X$ be any set and $B ⊂ P (X)$ be a family of subsets of $X$ that satisfy T1 and T2, then there is a unique topology in which B is an open base.
Where we call:
T1: Since $X ∈ T$, then $X$ will be a union of some elements of $B$, that is, $X =\bigcup_{λ∈Λ}B_λ$ where $B_λ ∈ B$. Since $X =\bigcup_{λ∈Λ}B_λ⊂\bigcup_{C∈B}C ⊂ X$, then have $\bigcup_{C∈B}B = X.$
T2: Let $B_1, B_2 ∈ B$ and $x ∈ B_1 ∩B_2$. Since B_1 and B_2 are open, then B_1 ∩B_2 is open and so there exists $B_3 ∈ B$ such that $x ∈ B_3 ⊂ B_1∩B_2.$

Answer 1

Define $\tau = \{\mathcal{U}\in \mathcal{P}(X)$ such that $\forall$ $x \in \mathcal{U}$ there exists some $V \in B$ such that $x \in V \subset \mathcal{U} \}$.

$\tau$ will be a topology of $X$, then try to prove the unicity.