let $(X,d)$ be a metric space and $(x_{n})$ a Cauchy sequence on $X$ ,
$A=\left \{ x_{n}:n=1,2... \right \}$ if ${A}'\neq \varnothing $ (where ${A}'$ is the set of the limit points of $A$) prove that there exist a $x\epsilon X : x_{n}\rightarrow x$
my solution :
let $ \epsilon >0$ and a $N_{0}\epsilon \mathbb{N}$
since $(x_{n})$ is Cauchy for every $n,m\geq N_{0}$ $d(x_{n},x_{m})<\varepsilon $ (1) and ${A}'\neq \varnothing \Rightarrow $ it exist a $x\epsilon X$ such that
$B(x,\varepsilon )\cap A\setminus \left \{ x \right \}\neq \varnothing \Leftrightarrow B(x,\varepsilon )\cap \left \{ x_{n} \right \}\neq \varnothing$ (2)
from (1) ,(2) can i say that for every $n\geq N_{0}$
$d(x,x_{n})<\varepsilon \Rightarrow x_{n}\rightarrow x$
i feel something is missing to complete 100% the proof.
There are things missing indeed. For instance, you did not tell us what $N_0$ is. Or what that $\varepsilon$ is.
You can do it as follows. Take $\varepsilon>0$. Since $x\in A'$, there is an increasing sequence $(n_k)_{k\in\Bbb N}$ of natural numbers such that $\lim_{k\to\infty}x_{n_k}=x$ and therefore there is some $K_0\in\Bbb N$ such that$$k\geqslant K_0\implies\left|x-x_{n_k}\right|<\frac\varepsilon2.$$And there is a $N_0\in\Bbb N$ such that$$m,n\geqslant N_0\implies|x_m-x_n|<\frac\varepsilon2.$$But then, if $n\geqslant\max\{n_{K_0},N_0\}$, you take some $k\in\Bbb N$ such that $k\geqslant K_0$ and that $n_k\geqslant N_0$ and then\begin{align}|x-x_n|&\leqslant|x-x_{n_k}|+|x_{n_k}-x_n|\\&<\frac\varepsilon2+\frac\varepsilon2\\&=\varepsilon.\end{align}