Intersection of topologies

The proof seems just fine.${}{}$

Here is how I understand it.

Condition 1) is trivially satisfied since X and empty set are in every topology they will be in the intersection as well.

Condition 2) and 3) Here we need to prove that arbitrary unions and finite intersections of the elements in the intersection will be in the intersection. We explain this using intersection of two topologies over $X$. Let $T_1$ and $T_2$ be two topologies on $X$. Let set $A$ and set $B$ are in the intersection of $T_1$ and $T_2$. Since $A$ and $B$ are in the intersection of $T_1$ and $T_2$ it implies that $A$ and $B$ are both in $T_1$ and $T_2$ considered separately. Since $T_1$ and $T_2$ are topologies and $A$ & $B$ are member of both of these topologies, both $T1$ and $T2$ will contain union (and intersection) of $A$ & $B$ (by definition of being topologies) . Thus the union (and intersection) of $A$ & $B$ will be present in the intersection of the topologies $T_1$ and $T_2$. Hence we proved both condition 2 and condition 3 of being a topology.

This argument can be extended to the intersection of all the families of topologies on X.