Let $\theta:\mathbb{N}\to\mathbb{N}$, with $\lim{\theta(j)}=\infty$, when $j\to\infty$. If $(x_j)$ is a Cauchy sequence in $M$, then $y_j=x_{\theta(j)}$, defines a Cauchy sequence in the metric space $M$.
My Approach: I think if $y_{j}=x_{\theta(j)}$, then if $\theta(j)=j$ where $\lim\theta(j)=\infty$, $y_{j}$ obvoiusly will be a cauchy sequence, but this is a particular case.
Fix $\epsilon > 0$. Since $\{x_j\}$ is Cauchy, $\exists~N > 0$ s.t. $\forall~n, m \geq N$, we have $d(x_n,x_m) < \epsilon$ where $d$ is the metric on $M$. Then since $\theta(j) \to \infty, \exists~N_0$ s.t. $\forall~l \geq N_0, \theta(l) \geq N$. Thus, for all $n',m' \geq N_0, \theta(n'),\theta(m') \geq N$, which implies $d(x_{\theta(n')},x_{\theta(m')}) < \epsilon$, i.e. $d(y_{n'},y_{m'}) < \epsilon$. Thus $\{y_n\}$ is Cauchy.
$(x_j)$ is a Cauchy sequence means that for any $\varepsilon>0$, there exists $N\in\Bbb N$ such that for $m,n>N$, we have $$|x_m-x_n|<\varepsilon/2.$$ Since $\theta(j)\to\infty$ as $j\to\infty$, there is a $j_0\in\Bbb N$ so that for all $j>j_0$, we have $\theta(j)>N$. In particular, for any $k,l>j_0$, we have $\theta(k)=m_0$ and $\theta(l)=n_0$ with $m_0,n_0>N$ by construction and $$|x_{\theta(k)}-x_{\theta(l)}|=|x_{m_0}-x_{n_0}|<\varepsilon/2<\varepsilon$$by the assumption above. Thus, it is a Cauchy sequence.
