Here is my Approach ....i am also getting (E) as the answer
let A be {1,2} (say apple 1, apple 2)
there is atleast one element in A : satisfied
for every element x in A there is a y in A such that S(x,y) : Since nothing is told about the symmetry of the relation, we can have S(1,2) and S(2,1) : so,satisfied
So as of now, A = {1,2} and S={(1,2),(2,1)}
now we can go throught the options :
(A) S(x,x) is not possible ....i dont have that in S ....so satisified .....and still finite A
(B) S(x,x) should be there......well, i will make S={(1,2)(2,1),(1,1),(2,2)}....satisfied and still finite A
(C) take the same case as above ....satisfied and still finite A
(D) Now i cant have (1,1) and (2,2)
but even with S={(1,2).(2,1)} this condition is satisfied ....still finite A
(E) i cant do this with (1,2) and (2,1) ...because the trasitivity makes it (1,1) which should not be there .....and whatever elements i add the transitivity will lead me to (x,x) .....because any element added to A should occur atleast once in the left side of an ordered pair.......so only solution is infinite A ...