Proposition Let $(M,d)$ be a metric space. If $K\subset M$ is compact, then $K$ is complete.
Proof Let $\{x_n\}_{n=1}^\infty \subset K$ be a Cauchy sequence, then $ \forall \varepsilon > 0 \quad \exists n_0 \in \mathbb N \quad \forall m,n\geq n_0 \quad d(x_n,x_m) < \varepsilon. $ $(\text{Step } 1)$ Since $K$ is compact, $\{x_n\}_{n=1}^\infty$ has an accumulation point $x^*\in K$ (already proved for every metric space). Let's see that $\lim_{n\to\infty} x_n = x^*$.
$(\text{Step } 2)$ Since $x^*$ is an accumulation point $\forall k\in \mathbb N \quad \exists x_{n_k} \quad\text{s.t.}\quad d(x_{n_k}, x^*) <\frac{1}{k} $
$(\text{Step } 3)$ Let $\varepsilon > 0$, then $\exists k_0 \quad n_{k_0} \geq n_0 \quad d(x_{n_{k_0}} x^*) < \varepsilon $. Now if $n \geq n_0$, then $ d(x_n,x*) \leq d(x_n, x_{n_{k_0}}) + d(x_{n_{k_0}}, x*) \leq \varepsilon + \varepsilon = 2\varepsilon $ so every Cauchy sequence converges and $K$ is complete.
Question Is this proof correct? I can't understand steps $(2)$ and $(3)$ (why uses $\frac{1}{k}$?).
Thanks in advance.
Edit Is this proof correct?
Let $\{x_n\}_{n=1}^\infty \subset K$ be a Cauchy sequence, by definition $\forall \varepsilon >0 \;\, \exists n_0\in \mathbb N \;\, \forall m,n \geq n_0 \;\, d(x_n, x_m) < \varepsilon$. Since $K$ is compact, the sequence has an accumulation point $x^*\in K$ (already proved). Let's see that $\lim_{n\to\infty}x_n = x^*$: since $x^*$ is an accumulation point, then exists a subsequence $\{x_{n_k}\}_{k=1}^{\infty}\subset \{x_{n}\}_{n=1}^{\infty}$ that converges to $x^*$, in other words \forall \varepsilon > 0 \;\, \exists k_0\in\mathbb N \;\, \forall k \geq k_0 \;\, d(x_{n_{k}}, x^*) < \varepsilon. Because of the triangular property $d(x_n,x^*) \leq d(x_n, x_{n_{k}}) + d(x_{n_{k}}, x^*) \leq 2\varepsilon$, so every Cauchy sequence converges and $K$ is complete.