The set of all equivalence classes from Cauchy sequences is complete Let $C(x)$ be the set of all Cauchy seq. on $X$ and define for $(x_n)_{n \in \mathbb{N}}, (y_n)_{n \in \mathbb{N}}$ the following relation
$$
(x_n)_{n \in \mathbb{N}} \sim (y_n)_{n \in \mathbb{N}} \Leftrightarrow d_X(x_n, y_n) \to 0 \text{ when } n \to \infty
$$
Let 
$$
\hat{X} = C(X)/\sim, \\ 
d_{\hat{X}}([(x_n)], [(y_n)]) = \lim_{n \to \infty} d(x_n, y_n)
$$
Show that $\hat{X}$ is complete
Let $(\gamma_m)_{m \in \mathbb{N}}$ be a Cauchy seq. in $\hat{X}$ i.e $\gamma_m = ([(x_n)]_m)$. How can I show that there exists $[(x_n)] \in \hat{X}$ such that for $m> n_0 \in \mathbb{N}$ $d_{\hat{X}}([(x_n)]_m, [(x_n)]) < \epsilon, \ \forall \epsilon$?

I'll assume you already proved the metric is well defined, indeed a metric, and that $X$ is dense in $\hat{X}$. Now, when you take the Cauchy sequence of points in $\hat{X}$, remember that not all the terms in it are in $X$. So choose seqeunces $(x_{nk})_{k\in \mathbb{N}}$ (of points in $X$) such that either $[(x_{nk})_{k\in \mathbb{N}}]=\hat{x}_n=$ or, if $\hat{x_n}\in X$, then choose $x_{nk}=\hat{x}_n$. Then, for a suitable choice of naturals $k_1 <k_2<\dotsc$, the sequence $(x_{nk_{n}})_{n\in\mathbb{N}}$ is Cauchy in $X$ and $(\hat{x}_n)$ will converge to its limit in $\hat{X}$.
Let me know if you'd like me to expand some details.
