Prove that for rational $x$,$y$ and $\epsilon$ if $|x-y| \leq \epsilon$ , $ \forall \epsilon > 0$ then $x=y$

Question

I know that this is a repeated question, but I wanted to show my attempt.

Suppose $x \neq y$ and (wlog) $x > y$, then $x$ can be written as

$x = y + \delta$, for some ($\delta > 0$ and $\delta \in \mathbb Q$)

Therefore $|x-y| = \delta$.

But since $\delta < \epsilon$ , $\forall \epsilon > 0, \epsilon \in \mathbb Q $ , and we assumed that $\delta > 0$ ; therefore contradiction arises. Hence $x=y$.

Is this a correct proof? I want to know what is missing or how to add more rigour in the last statement.

Answer 1

...But since $\delta < \epsilon$ , $\forall \epsilon > 0, \epsilon \in \mathbb Q $...

This part is not right.

Make sure to use the right argument with the right inequality!

Since $\delta >0$, there exists $\epsilon \in \Bbb Q$ such that $0 < \epsilon < \delta$ and therefore $\epsilon < |x-y| = \delta$, which is absurd.