$a$ and $b$ are fixed real numbers.
Claim: $a < b$ implies $a < b - \varepsilon$ for some $\varepsilon > 0$.
Proof: Utilise the fact that $a \implies b$ is equivalent to $b' \implies a'$.
So this is equivalent to proving $a \ge b - \varepsilon\,$ for all $\varepsilon > 0 \implies a \ge b$.
Now, as $a \ge b - \varepsilon$, take the infimum of both sides.
$\inf (a) \ge \inf \{b - \varepsilon \mid \varepsilon > 0\}$
$\inf (a) = a$ and the right-hand side $= b$, hence:
$a \ge b.$
Thus, $a \ge b - \varepsilon$ for all $\varepsilon > 0$ implies $a \ge b$. Which then proves our original claim.