I want to prove that a sequence is $d$-Cauchy iff $$\forall \epsilon>0, \exists N\in\mathbb N :\text{diam}(\{x_n:n\ge N\}) \le \epsilon.$$
Definition of $d$-Cauchy sequence in a metric space $(M,d)$:
A sequence $x_n$ is $d$-Cauchy in $(M,d)$ if $$\forall \epsilon > 0 \exists N\in\mathbb N \forall m,n\ge N\ (d(x_m,x_n) < \epsilon)$$
Note the strict inequality in the definition, because of which I've asked this question.
$[\Rightarrow]$ is pretty obvious. If $x_n$ is $d$-Cauchy, $$\forall \epsilon > 0 \exists N\in\mathbb N \forall m,n\ge N\ (d(x_m,x_n) < \epsilon)\implies \sup\{a,b: a,b\in \{x_n:n\ge N\}\} \le \epsilon$$
$[\Leftarrow]$ direction is what is troubling me because of the strict inequality.
How do I proceed?