R ∩ S is an equivalence relation, because in R ∩ S, we have all pairs of type (a,a) definitely, so R ∩ S is reflexive.
Moreover, if there is any pair (a,b) in R ∩ S, then that must have been present in both R and S, and thus (b,a) must have also been present in both R and S, and so in R ∩ S, so R ∩ S is also symmetric.
We can argue similarly for transitivity. If there are pairs (a,b) and (b,c) in R ∩ S, then they both must have been present in R as well as S, and so (a,c) would have also been present in both R and S, and so (a,c) must also be present in R ∩ S, hence R ∩ S is transitive also.
Now let us see R ∪ S. It is actually need not be equivalence. The intuition is we may violate the transitivity property because if some (a,b) comes from R, and some (b,c) comes from S, then we will not have (a,c) in R ∪ S (assuming we don't have (a,c) in R or S.
So let us construct a counter example for R∪S to be equivalence statement. Let A = {1,2,3}
Now let R = {(1,1),(2,2),(3,3),(1,2),(2,1)}, and let S = {(1,1),(2,2),(3,3),(2,3),(3,2)}. Now R ∪ S = {(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)}. This violates transitivity as we have (1,2) and (2,3), but not (1,3).
So option (C) is correct.