Proof that a quotient metric space is complete Let $(X,d)$ a metric space and $\sim$ a equivalence relation such that : 


*

*$\forall x\in X$ : $[x]=\{y\in X \vert y \sim x \}$ is closed.

*If $[x] \neq [y]$ : $d([x],[y])=d(a,[y]), \forall a\in[x]$
Define in $\dfrac{X}{\sim}$ : $D([x],[y])=d([x],[y])$ what is a metric (I've proved it). Prove that if $(X,d)$ is complete then $(\dfrac{X}{\sim},D)$ is complete. 
My attempt : 
Let $\{[x_n]\}$ a cauchy sequence of $\dfrac{X}{\sim}$. Then for $\epsilon >0$ exists $n_0 \in \mathbb{N}$ such that :
$$ \forall m,n \geq n_0  \implies  D([x_m],[x_n])<   \dfrac{\epsilon }{2}  \implies D([x_m],[x_{n_0}])< \dfrac{\epsilon }{2}$$
By the definition of $D$ exists $p_m \in [x_m] $ such that 
$$ d(p_m,x_{n_0})< \dfrac{\epsilon }{2} \implies d(p_m,p_n)<\epsilon  $$ 
$\{p_m\}_{m\geq n_0}$ is cauchy?
 A: Here it is a way to get a Cauchy sequence from the elements of the $[x_n]'$s.
There are two facts that we will use without proving:
(1) If a sequence $(x_n)_{n\in\mathbb N}$ satisfies $d(x_n,x_{n+1})\leq 2^{-n},\ \forall n\in\mathbb N$, then it is a Cauchy sequence.
(2) The converse of (1) is true in the following sense: any Cauchy $(x_n)_{n\in\mathbb N}$ admits a subsequence $(x_{n_k})_{k\in\mathbb N}$ satisfying $d(x_{n_k},x_{n_{k+1}})\leq 2^{-k},\ \forall k\in\mathbb N.$
If $([x_n])_{n\in\mathbb N}$ is Cauchy, by fact (2), we fix a subsequence $([x_{n_k}])_{k\in\mathbb N}$ such that
$$d([x_{n_k}],[x_{n_{k+1}}])\leq 2^{-k},\ \ \forall k\in\mathbb N.$$
Fix some $p_1\in[x_{n_1}]$, so
$$
d(p_1,[x_{n_2}]) = d([x_{n_1}],[x_{n_2}])\leq \frac{1}{2}.
$$
Then we may fix $p_2\in[x_{n_2}]$ such that 

$
d(p_1,p_2)\leq d(p_1,[x_{n_2}]) + \dfrac{1}{2} \leq 1.
$

To fix $p_3$, remember that $$d(p_2,[x_{n_3}])=d([x_{n_2}],[x_{n_3}])\leq \dfrac{1}{4},$$ so we may fix $p_3\in[x_{n_3}]$ such that 

$
d(p_2,p_3)\leq d(p_2,[x_{n_3}]) + \dfrac{1}{4} \leq \dfrac{1}{2}.
$

To fix $p_4$, remember that $$d(p_3,[x_{n_4}])=d([x_{n_3}],[x_{n_4}])\leq \dfrac{1}{8},$$ and we may fix $p_4\in[x_{n_4}]$ such that 

$
d(p_3,p_4)\leq d(p_3,[x_{n_4}]) + \frac{1}{8} \leq \frac{1}{4},
$

and so on.
This way we get a sequence $(p_k)_{k\in\mathbb N}$ such that $p_k\in[x_{n_k}],\ \forall k\in\mathbb N$ satisfying
$$
d(p_k,p_{k+1})\leq 2^{k-1},\ \forall k\in\mathbb N,
$$
and, using fact (1) we conclude that $(p_k)_{k\in\mathbb N}$ is a Cauchy sequence.
