Suppose that {xm} is a Cauchy sequence, and that xm(k) is a subsequence of {xm} which converges to some point p. Prove that {xm} itself must also converge to p.
Let $\displaystyle \epsilon>0 $
Since $\displaystyle \{x_n\} $ is Cauchy, we know for some $\displaystyle N>0 \;\; \forall\; n,m>N, \; d(x_n,x_m)<\tfrac{\epsilon}{2} $.
Since $\displaystyle \{x_{n_k}\} $ converges to $\displaystyle p $ we also know for some $\displaystyle M>0 \;\; \forall n_i,m>M, \; d(x_{n_i},p)<\tfrac{\epsilon}{2} \text{ and } d(x_{n_i},x_m)<\tfrac{\epsilon}{2} $.
By the triangle inequality $\displaystyle d(x_m,p)<\epsilon $.