To prove two sets are equal, you have to show they are both subsets of one another.
• X ∩ (⋃ ) = ⋃ S ∈
Let x ∈ X ∩ (⋃ ). Then x ∈ X and x ∈ ⋃ . The latter means that x ∈ S for an arbitrary set S ∈ . So x ∈ X and x ∈ S, meaning x ∈ X ∩ S. That is enough to say that x ∈ ⋃ X ∩ S . So X ∩ (⋃ ) ⊆ ⋃ X ∩ S .
For the other direction, the proof is essentially the reverse. Let x ∈ ⋃ X ∩ S . Then x ∈ X ∩ S for some S ∈ , so that x ∈ X and x ∈ S. Because x ∈ S and S ∈ , we have that x ∈ ⋃ , and so x ∈ X ∩ (⋃ ). So ⋃ S ∈ ⊆ X ∩ (⋃ ).
QED
• X ∪ (⋂ ) = ⋂ X ∪ S
Let x ∈ X ∪ (⋂ ). Then x ∈ X or x ∈ ⋂ . If x ∈ X, we're done because that would guarantee x ∈ X ∪ S for any set S, and hence x would belong to the intersection. If x ∈ ⋂ , then x ∈ S for all S ∈ , so that x ∈ X ∪ S for all S, and hence x is in the intersection. Therefore X ∪ (⋂ ) ⊆ ⋂ X ∪ S .
For the opposite direction, let x ∈ ⋂ X ∪ S . Then x ∈ X ∪ S for all S ∈ . So x ∈ X or x ∈ S for all S. If x ∈ X, we're done. If x ∈ S for all S ∈ , then x ∈ ⋂ , and we're done. So ⋂ S ∈ ⊆ X ∪ (⋂ ).
QED