Let $(X, d)$ be a compact metric space. Then, $(X, d)$ is both complete and bounded.
The author suggest to prove completeness and boundedness separately, using contradiction.
Suppose that $(X, d)$ is not bounded. Then, there exists no $B_X(x, r)$ which contain $X$. That is, for any $x \in X$, we can find $x_n \in X$ such that $d(x, x_n) >n$ for every $n \in \mathbb{N}$. So $(x_n)_{n=1}^\infty$ diverges and all its subsequence must diverge too.
Suppose that $(X, d)$ is not complete. Then, there exists a Cauchy sequence $(x_n)_{n=1}^\infty$, which is not convergent in $X$. That is, there exists $\epsilon>0$ such that for every $n$, $d(x, x_n) > \epsilon$, where $x$ is any element in $X$. This implies that all its subsequence does not converge in $X$. However, it is possible that $(x_n)_{n=1}^\infty$ converges to $x \not\in X$. That is, $x$ might be an adherent point of $X$, which is not contained in $X$ (i.e., for every $r>0$, $B_X(x, r) \cap X \not= \emptyset$). I think that in this case, every its subsequence should converge to $x$. What is wrong in my proof? Any help would be appreciated.
$\exists \epsilon >0 \,\forall x\in X \,\forall n\,(d(x,x_n)>\epsilon)$ implies that no $x_n$ is equal to any member of $X,$ which is absurd.
Suppose $S=(x_n)_n$ is Cauchy and has a convergent sub-sequence. Then $S$ is convergent. This is verbatim to the special case of sequences in $\Bbb R$ except we write $d(x,y)$ instead of $|x-y|$:
$\quad$ Suppose $f:\Bbb N\to \Bbb N$ is strictly increasing and $(x_{f(n)})_n$ converges to $x.$ Given $r>0,$ take $n_1 \in \Bbb N$ such that $n>n'\ge n_1\implies d(x_{n'},x_n)<r/2$ (because $S$ is Cauchy). Now there exists $n_2$ with $f(n_2)\ge n_1$ and $d(x,x_{f(n_2)})<r/2.$ Then $$n\ge f(n_2)\implies d(x,x_n)\le d(x,x_{f(n_2)})+d(x_{f(n_2)},x_n)<$$ $$<r/2+d(x_{f(n_2)},x_n) $$ $$<r/2+r/2 \quad \text { (as $f(n_2)$ and $n$ are $\ge n_1$)}.$$