Proving $∀ε>0:x<y+\epsilon ⇒ x<y$

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 $x < y$.

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 $x=y+\varepsilon$.