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.