There is another way of looking at this problem which uses first-order logic concepts. Recall the truth table of A implies B, or A => B.
The definition of reflexivity is that if element "a" belongs to A, then (a, a) must belong to S. This is an implication whose left side is true and right side is false, for all elements in A. Therefore the result is false.
For symmetry, for a pair of elements a,b in A, if (a, b) belongs to S, then (b, a) must also belong to S. The left side of this implication is false, so it does not matter if the right side is true or false, the result will always be true. Thus S is symmetric.
Similarly, for transitivity, given elements a,b,c in A, if (a, b) belongs to S and (b, c) belongs to S, then (a, c) must belong to S. The left side is once again false, so the result of the implication is true. S is transitive.
The same reasoning can be extended to anti-symmetry although it is not required here. Therefore the answer is (D).