Let $\{x_n\}$ be a Cauchy sequence of rational numbers that do not tend to zero, showing that $\{\frac{1}{x_n}\}$ is a Cauchy sequence. I was not able to use the fact that $x_n$ does not tend to 0, so I am not entirely sure whether or not my proof is correct, can someone check for me? thanks
Proof.
Suppse $\{x_n\}$ is a Cauchy sequence, that do not tend to $0$, Claim then there exists $M>0$ such that $|x_n|>M$ for all $n>0$,
Proof of Claim, suppose $x_n>0$ for all $n$, pick $\epsilon=C>0$,where $C<x_{N+1}$ then we have $n>N\implies x_n>x_{N+1}-C>0$, now pick $M=\min\{x_1,\dots , x_{N},x_{N+1}-C\}$ then we have $x_n>C>0$ for all $n$. Similarly done for $x_n<0$ and hence $|x_n|>M>0$ for all $n$.
Now, since $M<|x_n|\implies \frac{1}{|x_n|}<M$
Let $|x_n-x_m| <\frac{\epsilon}{M^2}$, where $n,m>N$ hence we have $|\frac{1}{x_n}-\frac{1}{x_m}|=\frac{|x_n-x_m|}{|x_nx_m|}<\frac{M^2\epsilon}{M^2}=\epsilon$ and hence $\{\frac{1}{x_n}\}$ is Cauchy Sequence.