I would have been inclined to do an indirect proof:
Suppose a/b < (a+x)/(b+y) < x/y is NOT true. Then either a/b>= (a+ x)/(b+y) or (a+ x)/(b+y)>= x/y.
If a/b>= (a+ x)/(b+ y) then a(b+y)>= b(a+ x) which is the same as ab+ ay>= ab+ bx and so ay>= bx which (assuming b and y are non-zero which is implied by the hypothesis) a/b>= x/y, contradicting the hypothesis.
Similarly, if (a+ x)/(b+ y)>= x/y, then y(a+ x)>= x(b+y) which is the same as ay+ xy>= bx+ xy and so ay>= bx again which gives the same contradiction.