Every topology in your family must contain the empty set, so the intersection of the topologies will also contain the empty set. Ditto for X itself. So the intersection contains the empty set and the full set.
If the intersection of these topologies contains some family U of sets, then of course every topology in {Ta} contains those sets of U. Since these are topologies, and are closed under set union, so each of these topologies contains the union of U. Therefore, the union of U belongs to every topology in {Ta}, and hence lies in their intersection. So the intersection of Ta is closed under arbitrary union.
The argument for the final requirement, finite intersection, is similar. Try it for yourself.
