Proof that a quotient metric space is complete

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

1. $$\forall x\in X$$ : $$[x]=\{y\in X \vert y \sim x \}$$ is closed.
2. 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?

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.