Show that a Cauchy sequence has a fast-Cauchy subsequence

Just select $x_{n_{k}}$ ($k=1,2,\ldots$) according to whether $d(x_{n_{k+1}},x_{n_{k}})<2^{-k}$. This is always possible since you can take $\epsilon_{k}=2^{-k}$ in the Cauchy Criterion.

Let $\mathcal{Y}=\{y_k\}_{k=0}^\infty$ be your Cauchy sequence. Choose any $0<r<1$. Then, since $\mathcal{Y}$ is Cauchy, for every power $r^k,\,k\in\mathbb{N}$ $\exists n(k)\in\mathbb{N}$ such that $|y_j -y_{j+1}|<r^k=\epsilon_k,\,\forall j>n(k)$. So now

$$0<\sum_{j=0}^p|y_j -y_{j+1}| <\sum_{j=0}^p r^k$$

for any positive $p$. The geometric series on the right converges, whence the claim is proven.