If $x - ε ≤ y$ for all $ε>0$ then $x ≤ y$ I've been asked to prove the following, 
if $x - ε ≤ y$  for all $ε>0$ then $x ≤ y$.
I tried proof by contrapositive, but I keep having trouble choosing the right $ε$. Can you guys help me out?  

Suppose $x > y \implies \epsilon = x-y > 0 \implies x = y + \epsilon > y + \dfrac{\epsilon}{2}$, contradiction.
