Let $\{x_n\}$ be a sequence of Rational numbers: $$\exists \ \ 0 < \epsilon \in \Bbb Q \ \ \exists \ \ N \in \Bbb N: \lvert x_n - x_m \rvert < \epsilon \ \ \forall \ \ n,m \geq N$$
This is not guaranteed to be a Cauchy Sequence since $\epsilon$ is given. Is it true that all terms $x_n$ for $n \geq N$ are within a band of width $\epsilon$, that is, $$\exists \ \ a \in \Bbb Q: \ \ \forall \ \ n \geq N, \ \ a \leq x_n \leq a + \epsilon$$
Drawing some diagrams, it seems possible, but I have not been able to prove or disprove this.
Attempt: Suppose $$\forall \ \ a \in \Bbb Q: \exists \ \ n \geq N: a > x_n \vee x_n > a + \epsilon$$ Choose $a=x_N$. Then $n \geq N \implies x_n < x_N +\epsilon$. So $x_n < x_N$. I do not know how to proceed from here.
EDIT: It is not necessary that $a$ be one of the terms in the sequence, just some rational number.