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.

Suppose $T$ is a topology stronger than every topology in the family ${T_i}$.  Let $T'$ be the topology generated by the open subbase $\cup_i T_i$.  You need to prove that $T$ contains  $T'$.  If this is true then...
Edit: changed "$T$ is stronger than  $T'$" to "$T$ contains  $T'$". This should be more clear.
