You cannot prove it. It isn't true!
Think about what R(A) and R(B) mean. The first is a set of ordered pairs of objects in A, the second is a set of ordered pairs in B. Example: A= {a, b} so one possible relation on A is "pRq if and only if both p and q are A". That is R(A)= {(a, a), (a, b), (b, a), (b, b)}. And if B= {x, y} then one possible relation is {(x, x), (x, y), (y, x), (y, y)}.
R(A)U R(B) is the union of those two sets- it contains {(a, a), (a, b), (b, a), (b, b), (x, x), (x, y), (y, x), (y, y)}.
But the equivalent R on A U B contains all pairs of elements of both A and B. In particular it contains (x, a) and (b, y) which R(A) U R(B) does not.
Actually, the problem as stated, doesn't makes sense. You have three different sets, A, B, and A U B. Since a relation on a set is defined as a "subset of ordered pairs of elements from the set", you cannot have the same relation on all three. Unless A and B are subsets of some super-set and R is explicitely defined on that set. If so, you have not told us that.
I recommend you go back and read over this problem more carefully.
Please please please help me if you know how