Proving $∀ε>0:x
If $x, y \in \Bbb R$ and $x<y+\epsilon$ for every $\epsilon >0$, then $x<y$. 
Okay so I went about this by proving the contrapositive.
Proof: Let $x,y\in\Bbb R$ and let $\epsilon>0$. Suppose $x\ge y$. Take $\epsilon = x-y$. This implies that $x=y+\epsilon$, as needed.   
Is this a valid proof or not? 
 A: The statement you want to prove is not true. 
Take $x = y =0$. You have, for all $\varepsilon > 0$, $x < y + \varepsilon$ (because $\varepsilon > 0$), but of course you don't have $x < y$.
A: The proof is invalid for the simple fact that the statement is false, so you cannot prove it.
The correct statement is “if, for all $\varepsilon>0$, $x<y+\varepsilon$, then $x\le y$”.
Now your proof works! Suppose $x>y$ (that is, “not $x\le y$”) and take $\varepsilon=x-y$; then $\varepsilon>0$ and $x=y+\varepsilon$.
A: 1) The statement is false.
Counter example:  Let $x = y$.  Then $x = y < y+\epsilon$ for all $\epsilon > 0$ but $x \not < y$.
2)  Your error is in claiming setting $\epsilon = x - y$.  If $x =y$ then that $\epsilon \not > 0$.
Your proof is doomed to fail for $x = y$ because the statement is false for $x =y$.
3) But you have proven that $x>y$ is impossible. That would mean if $\epsilon = x - y > 0$ then $x = y + \epsilon$ which contradicts our hypothesis that $x < y + \epsilon$ for all $\epsilon>0$.
4) So a TRUE statement would be: if $x < y + \epsilon$ for all $\epsilon > 0$ then $x \le y$ and you successfully have proven that.
