Discrete Mathematics: $x\leq y+\epsilon \implies x\leq y$ 
Let $x$ and $y$ be real numbers. Prove that if $x\leq y + \epsilon$ for every positive real number $\epsilon$, then $x\leq y$.

I would like a hint as to how to prove this. Thank you. Pictorial proof would be nice too.

So this is how I word-smithed the answer given by the author of my accepted answer:
We will proceed by showing the contrapositve. We are given that
$$x\leq y+\epsilon \implies x\leq y,$$
so the contrapositive is resolved as
$$x > y \implies x>y+\epsilon,$$
or by substituting $x-y = \omega$ simply
$$\omega > 0 \implies \omega > \epsilon,$$
but if $\omega >0$, then $\epsilon = \frac{\omega}{2} > 0$; however, $\omega \leq \frac{\omega}{2}$ is false. Therefore, we can assert that for all $\epsilon > 0$
$$\omega\leq\epsilon \implies \omega \leq 0;$$
quod erat demonstrandum.
Here is my pictorial representation:

 A: The claim is equivalent to showing that if $\omega\leq \epsilon$ for each $\epsilon >0$, then $\omega\leq 0$.
But, if $\omega>0$, then $\epsilon=\frac \omega 2 >0$ and $\omega \leq \frac\omega 2$ does not hold. Having proven the contrapositive, we can assert that $$(\forall\epsilon >0\;;\;\omega\leq\epsilon )\implies \omega \leq 0$$
Now let $\omega =x-y$.
Pictorially If for any $\color{green}{\epsilon >0}$ we choose, $\omega$ is to the left of  $\color{green}{\epsilon}$, it must be the case $\omega$ is to the left of the green bar, that is, on the red side $\color{red}{\omega <0}$ (strictly negative numbers) or that it is on the breaking point, that is $\color{orange}{\omega=0}$.

ADD The contrapositive of the assertion is $$\omega >0\implies (\exists \epsilon >0:\omega\not\leq \epsilon)$$ or, which is the same, $$\omega >0\implies (\exists \epsilon >0:\omega> \epsilon)$$
We proved the contrapositive with $\epsilon =\omega /2$.
A: Consider $x,y$ be universally quantified, then
$$\forall \epsilon\ (x\leq y+\epsilon) \implies x\leq y$$
$$\neg (x\leq y)\implies \neg\forall \epsilon\ (x\leq y+\epsilon)$$
$$\neg (x\leq y)\implies \exists \epsilon\ \neg(x\leq y+\epsilon)$$
$$x>y\implies \exists \epsilon\ (x>y+\epsilon)$$
For $x>y$, we know $x>\tfrac{x+y}{2}=y+\tfrac{x-y}{2}$ and $\tfrac{x-y}{2}>0$, so it fits the bill.
