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?  
 A: First of all, understand that most of the problems dealing with $\epsilon, \delta$ are an exercise in inequalities and use very very elementary ideas. The difficulty of such problems arises mainly because of use Greek letters and partially because students don't really understand inequalities as much as they understand $+, -, \times, /, =$.
You wish to start by contrapositive which is almost similar to method of contradiction which we use in this answer. Let us then assume that $x > y$ i.e. $x - y > 0$. Now our goal is to use the hypotheses $x - y > 0$ to contradict the following statement $$\text{for all }\epsilon > 0, x - \epsilon \leq y$$ One way to contradict this statement is to find an $\epsilon > 0$ such that $x - \epsilon > y$ i.e. $\epsilon < x - y$.
Clearly this is possible because of the hypotheses that $x - y > 0$ and we just need to choose a positive $\epsilon$ smaller than positive number $x - y$. We don't specifically need to have an explicit expression for $\epsilon$ like $\epsilon = (x - y)/2$ (doing this always looks magical to a beginner). What is really needed is to understand that by hypotheses $A = x - y$ is positive and any positive $\epsilon$ less that this specific positive $A$ will do our job.
So after all this understanding given in last few paragraphs you just write the answer like this:

Suppose on the contrary that $x > y$ so that $x - y$ is positive and hence there is a positive $\epsilon $ which is smaller than $x - y$ so that $0 < \epsilon < x - y$ and therefore $x - \epsilon > y$ which contradicts the fact that $x - \epsilon \leq y$ for all $\epsilon > 0$. Therefore our supposition is wrong and we must have $x \leq y$.

In the above answer the fundamental idea used is the fact that given any positive number, there exists a smaller positive number. This is a fact of inequalities which is so obvious yet students find it difficult to use in problems such as these because of the use of Greek letters.
A: Hint
$$
x-\varepsilon \leq y\iff  x-y \leq \varepsilon
$$
Suppose that $x>y$. This implies that $\bar \varepsilon=\frac {x-y}{2}>0$, then $\dots$

The contrapositive of $$\forall \varepsilon > 0 \, (x-\varepsilon \leq y) \rightarrow x\leq y$$ 
Is
 $$x> y \rightarrow \exists \varepsilon>0\,( x-\varepsilon > y)$$
Suppose that for all $\varepsilon>0, $ we have that $x-\varepsilon \leq y$, then $\frac {x-y}{2}$ is positive, then $\dots$ 
A: Suppose $x > y \implies \epsilon = x-y > 0 \implies x = y + \epsilon > y + \dfrac{\epsilon}{2}$, contradiction.
A: 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.
