So far I've only proven this fact given both $\{x_n\},\ \{x_m\}$ are either positive or negative:
Let $\varepsilon >0$. I shall prove that given $\varepsilon = 2|q|^N$, the succession is bounded by $\varepsilon$.
We know $|q|<1$. Then $|q|\cdot|q|^n = |q|^{n+1}<|q|^n \implies \{x_n\}>\{x_{n+1}\}$, so $\{x_n\}$ is decreasing for $\{x_n\} > 0$.
Now, if $N = \min\{m, n\}$, then $|q|^n+|q|^m < 2|q|^N$. We can prove that if a, b are either both positive or both negative, then $|a-b|\leq |a+b|$.
Substituting $a = q^n,\ b=q^m$ we have that $|q^n- q^m|\leq|q^n+ q^m|\leq |q^n|+ |q^m|< \varepsilon= 2|q|^N$, Then, $|q^n- q^m|<\varepsilon$ and it's a Cauchy succession.
My problem here is that I have no idea what to do if $\{x_m\}<0$ and $\{x_n\}>0$. Is it even solvable this way?
Thank you.
You can easily show that if $0\le q<1$ then $q^n\to 0.$ (In fact, it's probably much easier to show this at the outset, and then the sequence is $\textit{automatically}$ Cauchy.)
Now, if $-1<q\le 0$ then $q^n=(-1)^n|q|^n$ and since you've shown that $|q|^n$ is Cauchy, it follows from the fact that $-|q|^n\le (-1)^n|q|^n\le |q|^n$ and $|q|^n\to 0,$ that $(-1)^n|q|^n$ is Cauchy.