Let $(X,d)$ be a compact metric space. Then $(X,d)$ is both complete and bounded.
My solution
The space $(X,d)$ is indeed complete. This is because every Cauchy sequence which admits a convergent subsequence is also convergent.
Now it remains to prove the bounded part.
Let us suppose otherwise that $X$ is unbounded.
Thus for every $r > 0$ there is an element $x\in X$ such that $x\not\in B(x_{0},r)$.
In particular, if we choose $r = n$, there corresponds $x_{n}\in X$ such that $d(x_{n},x_{0}) \geq n$.
But, since $x_{n}$ admits a subsequence $x_{f(n)}$ which converges, such subsequence is bounded: $d(x_{f(n)},x_{0}) < N$ for some natural $N$.
Thus if we choose $n_{0} = N + 1$ and noticing that $f(n_{0}) \geq n_{0}$, we conclude that \begin{align*} N+1 = n_{0} \leq f(n_{0}) \leq d(x_{f(n_{0})},x_{0}) \leq N \end{align*} which is a contradiction. Therefore $X$ is bounded.
Could someone please point out any theoretical flaw, if there is one?
The subsequence converges to some point $y_0$ and we get $d(x_{f(n)},y_0) <N$ instead of $d(x_{f(n)},x_0) <N$. (Though $d(x_{f(n)},x_0) $ is also bounded you have give some argument to justify this).