This should follow from the fact that $x>y$ if and only if $x-y>0$.
Because we are saying there exists an $a \in S$ such that this holds true, my thought was to do a proof by contradiction (supposing it holds true for all $a$ and finding one example where it doesn't). However, I'm having a hard time negating this statement. Right now I have:
If $y>x$, then for all $a$ such that $a \le 0$, $x+a\ne y$.
Which seems to be true. I may be barking up the wrong tree here, help approaching this proof would be much appreciated.