What I am thinking so far is following:
Construct another sequence $\{b_n\}$ such that $b_1=x_2, b_2=x_3, \ldots, b_n=x_{n+1}.$ Since $\{x_n\}$ is a Cauchy sequence, $\{b_n\}$, as a subsequence of $\{x_n\},$ is therefore also a Cauchy sequence.
Since Cauchy sequences are bounded, we can find $M_1, M_2 \in Q$ such that $|x_n| \le M_1$ and $|b_n| \le M_2$ for all $n.$ Let $M = \max\{M_1, M_2\}.$
Since $\{x_n\}$ is a Cauchy sequence, we can find $N_1$ such that for all $m,n \ge N_1,$ $|x_n-x_m| < \frac \varepsilon{2M}$. Similarly, we can find $N_2$ such that for all $m,n \ge N_2,$ $|b_n-b_m| < \frac \varepsilon{2M}.$
Let $N = \max\{N_1, N_2\}$ and $n,m \ge N.$
Therefore, $$|y_n-y_m|=|x_n x_{n+1}-x_m x_{m+1}|=|x_n b_n-x_m b_m|=|x_n b_n-x_m b_n+x_m b_n-x_m b_m|=|b_n||x_n-x_m|+|x_m||b_n-b_m| \le M\times\frac \varepsilon{2M} + M\times\frac \varepsilon{2M} = \varepsilon$$
QDE.
I wonder if there is any flaw in this proof and if there is a better way to prove it. Thank you!