Let $T: X \rightarrow Y$ be a continuous linear operator. $(X,\rho), (Y,\xi)$ are linear metric spaces and $\{x_n\} \subset X$ is a Cauchy sequence.
I need to show that $\{Tx_n\}$ is a Cauchy sequence.
For $T$ to be continuous means:
$\forall \varepsilon>0\,\, \exists \delta>0$ such that $\rho(x,y)<\delta \implies \xi(Tx,Ty)<\varepsilon.$
And for sequence to be Cauchy means:
$\rho(x_n,x_m)\rightarrow 0$ for $m,n \rightarrow \infty.$
How do I show that $\xi(Tx_n,Tx_m)\rightarrow 0$ for $m,n \rightarrow \infty?$ So far I suspect that the operator needs to be linear, because there are counterexamples for nonlinear continuous mappings (Namely $\{\frac{1}{n}\}$ and $T(x)=\frac{1}{x}$.).
I'm sorry if I wasn't thorough enough with my search but, I've only seen a similar question here with normed vector spaces instead of metric or the mapping was uniformly continuous and I wasn't able to translate it to my problem.
EDIT: Forgot to include this attribute of metrics: $\rho(x_1+x,x_2+x)=\rho(x_1,x_2) \forall x_1,x_2 \in X$