Let $(x^{(n)})_{n=m}^{\infty}$ be a Cauchy sequence in $(X,d)$. Suppose that there is some subsequence $(x^{n_{j}})_{j=1}^{\infty}$ 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
Let us start with the definitions. The sequence $x_{n}$ is Cauchy iff for every $\varepsilon/2 > 0$ there is a natural number $N_{1}\geq m$ such that \begin{align*} k\geq j\geq N_{1} \Rightarrow d(x_{j},x_{k}) \leq \varepsilon/2 \end{align*}
On the other hand, if its subsequence $x_{f(n)}$ converges to $x_{0}$, then for every $\varepsilon/2 > 0$, there is a natural number $N_{2}\geq m$ such that \begin{align*} f(n)\geq N_{2}\Rightarrow d(x_{f(n)},x_{0})\leq\varepsilon/2 \end{align*}
Then we have to prove that for every $\varepsilon > 0$ there is a natural number $N\geq m$ such that \begin{align*} n\geq N \Rightarrow d(x_{n},x_{0})\leq\varepsilon \end{align*}
Indeed, this is the case due to the triangle inequality.
Precisely, since $f(n)\geq n$, for every $\varepsilon > 0$ there is a natural number $N = \max\{N_{1},N_{2}\}$ 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}) \leq \varepsilon \end{align*}
and $x_{n}\to x_{0}$.
Am I making any conceptual mistake? Any improvement to my answer is appreciated.