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?  

For shits and giggles and the learning experience it brings, here's a proof intended to reinforce your concepts of sup/inf.
$x - \epsilon \le y \forall \epsilon >0$
$x - y \le \epsilon \forall \epsilon > 0$
So $x-y$ is a lower bound for $\{\epsilon > 0\} = (0,\infty) $.
So $x-y \le \inf (0,\infty) $.
It's good practise to see if you can prove $\inf (0,\infty) =0$.  I'm not going to do it here.  Let's assume you've seen it done.
$x-y \le \inf (0,\infty) = 0$.
So $x \le y $.
We're done.
Okay the contra positive proof is easier, more direct and better.  But analysis classes are going to expect us to get up to speed on bound proofs very quickly and with little practise.  
We might as well get comfortable with them.
