About lower upper bound of a subset of all topologies for a non empty set X 
How i may prove what is stated? That the topology generated using the theorem D as in the image, is equal to the lower upper bound of a non-empty family of topologies for a non empty set $X$. 
Definitions: 
   The lower upper bound of a non-empty family of topologies of $X$ is the topology whose arise from the intersection of all topologies stronger than each topology in the family. 
A topology T is stronger than R if R is cointained in T.
I tried to prove that one set is cointained in the other by proving that every open set from one is in the another, but i could not, in special because the union of the topologies. One there know how to prove it or have a hint? Thank you very much.

If you want to prove it in detail, all you have to show is that last statement:

the class of all unions of finite intersections forms a topology

which is to say, this class is closed under the formation of finite intersection and arbitrary unions.  
Doesn't it follow by construction?
(Notice that $\bigcup \varnothing = \varnothing$ and $\bigcap \varnothing = X$.)
