Sequentially compact metric space is complete and totally bounded [duplicate]

Just to cover both results:

Assume that $(x_n)_{n\in\mathbb{N}}$ is a Cauchy sequence in a sequentially compact space $X$. Introduce a convergent subsequence $(x_{n_k})_{k\in\mathbb{N}}$ of $(x_n)$ with $x_{n_k}\to x.$ Let $\epsilon>0$ be given. Choose $N$ such that $\rho(x_i,x_j)<\epsilon/2,~i,j\geq N$. Choose $n_k>N$ such that $\rho(x_{n_k},x)<\epsilon/2$. Then we have \begin{equation*} \rho(x,x_N)\leq\rho(x,x_{n_k})+\rho(x_{n_k},x_N)<\epsilon/2+\epsilon/2=\epsilon \end{equation*} proving completeness.

Assume by contradiction that $X$ is not totally bounded. Take $\epsilon>0$ such that $X$ cannot be covered by a collection of finitely many $\epsilon -$balls. Choose $x_1\in X,~x_2\in X\setminus B_{\epsilon}(x_1),~x_3\in X\setminus B_{\epsilon}(x_1)\setminus B_{\epsilon}(x_2),...~$ .Then we have a sequence $(x_n)$ which cannot contain a convergent subsequence (since $\rho(x_i,x_j)\geq \epsilon~\forall i\neq j$). $~_{\square}$

Let $\epsilon>0$. If $X$ is finite the statement is obvious. If not, take some $x_1\in X$, then take $x_2\in X\setminus B(x_1,\epsilon)$,..., then take $x_{n+1}\in X \setminus\bigcup\limits_{k=1}^n B(x_k,\epsilon)$... (if you eventually run out of points you have proven the statement).

Now use the fact that $\{x_n\}$ contains a subsequence that converges.

Suppose it is not bounded. Then you can, asuming $X$ is not empty (nor finite) so $x_0 \in X$, you construct this sequence : $x_1 \in B(x_0,1)$, $x_2 \in B(x_0,2)-B(x_0,1)$,... so on. Has this sequence a convergent subsequence?