If $|A-B| < \epsilon$ for every $\epsilon > 0$ where $A, B$ are numbers, then $A = B$.

Can you please check my proof of this statement using proof by contradiction?

Claim:

If $|A-B| < \epsilon$ for every $\epsilon > 0$ where $A, B$ are numbers, then $A = B$.

Proof:

Assume $A \neq B$.

Let $|A - B| = m$.

$\implies$ Since we can take any real number greater than $0$ for $\epsilon$, we can always find $\epsilon$ such that $\epsilon \le m$.

$\implies$ Pick $\epsilon$ such that $\epsilon \le m$.

$\implies$$|A - B| = m \ge \epsilon$.

Contradiction.

$\implies$$A = B$.

Q.E.D.