Let $(x_{n})_{n=m}^{\infty}$ be a Cauchy sequence in $(X,d)$. Suppose that there is some subsequence $(x_{f(n)})_{n=m}^{\infty}$ of this sequence which converges to a limit $x_{0}\in X$. Then the original sequence $(x_{n})_{n=m}^{\infty}$ also converges to $x_{0}$.
My solution
To begin with, notice that \begin{align*} d(x_{n},x_{0}) \leq d(x_{n},x_{f(n)}) + d(x_{f(n)},x_{0}) \end{align*}
Let $\varepsilon/2 > 0$. Since $x_{n}$ is Cauchy, there is a natural number $N_{1}\geq m$ such that \begin{align*} n'\geq n \geq N_{1} \Rightarrow d(x_{n'},x_{n}) < \varepsilon/2 \end{align*}
Once $f(n)$ is strictly increasing, we have that $f(n)\geq n$. Thus for every $n\geq N_{1}$, we have that \begin{align*} f(n)\geq n \geq N_{1} \Rightarrow d(x_{n},x_{f(n)}) < \varepsilon/2 \end{align*}
But we do also know that $x_{f(n)}\to x_{0}$. Consequently, there corresponds a natural $N_{2}\geq m$ such that \begin{align*} n\geq N_{2} \Rightarrow d(x_{f(n)},x_{0}) < \varepsilon/2 \end{align*}
Gathering both results, for every $\varepsilon > 0$ there corresponds a $N = \max\{N_{1},N_{2}\}$ such that \begin{align*} n\geq N \Rightarrow d(x_{n},x_{0}) \leq d(x_{n},x_{f(n)}) + d(x_{f(n)},x_{0}) < \varepsilon \end{align*} and we are done.
I am mainly concerned with the wording of the proof. Can someone point out any flaw?