$(x_n)$ be bounded, and $A=\{ x_n\ |\ n\in \mathbb{N}\}.$ If $A$ has only one accumulation point, say x, then $x_n$ is convergent to x. I was trying to prove the following proposition;

Let $(x_n)$ be a bounded sequence, and define $$A = \{ x_n\ |\ n\in
\mathbb{N}\}.$$ If $A$ has only one accumulation point, say $x$, then
  $x_n$ is convergent, and it is convergent to $x$.

However, assuming that $A$ has only one accumulation point is not useful, and if I consider the contrapositive statement, it is much more thing to prove I think, so I basically stuck.
So, how can we prove this proposition ?
Note: Any hint also would be appreciated.
Edit:
After @Carlos's answer, the sequence might be Cauchy instead of just being bounded.
 A: But, the result is not true. Consider the next sequence
\begin{equation}
x_n=
\left\{
\begin{array}{ccc}
\displaystyle\frac{1}{n} & \text{if} & n \ \text{is odd}\\\\
2 & \text{if} & n \ \text{is even}
\end{array}
\right.
\end{equation}
The sequence is bounded and only have one accumulation point (0 is accumulation point) but not converges.
EDIT: After the edit of the question, if the sequence is Cauchy, obviously the counterexample doesn't works. Then, let $\varepsilon>0$. Because $x_0$ is the only accumulation point, then, $\left(x_0-\frac{\varepsilon}{2},x_0+\frac{\varepsilon}{2}\right)\cap A\neq\emptyset$. But, the sequence is Cauchy, then, for the same $\varepsilon$, there exists $N\in\mathbb{N}$ such that for all $m,n\geq N$, $d(x_n,x_m)<\frac{\varepsilon}{4}$. Then, because the set $\left(x_0-\frac{\varepsilon}{2},x_0+\frac{\varepsilon}{2}\right)\cap A$ is infinite (why?), we can take $x_j\in \left(x_0-\frac{\varepsilon}{2},x_0+\frac{\varepsilon}{2}\right)\cap A$ such that $j\geq N$, then, if $n\geq N$, we have the next inequalities: $$d(x_n,x_0)\leq d(x_n,x_j)+d(x_j,x_0)<\frac{\varepsilon}{4}+\displaystyle\frac{\varepsilon}{2}<\varepsilon$$Thus, we have the desired result.
A: 1)$P$ is an accumulation point of a sequence $(x_n)_{n \in.\mathbb{N}}$ if there is a subsequence converging to $P.$
Let $(x_n)$ be bounded and consider the set
$A= ${$x_n |  n \in \mathbb{N}$}. Let $A$ have only one accumulation point $P$.
Since $P$ is accumulation point of A:
In every $\epsilon$ neighbourhood of $P$ there is an $x_n \not= P$.
$\rightarrow$
A subsequence of $(x_n)$ converges to $P.$
However, $(x_n)$ need not converge.
Example by CJimenez.
1) A subsequence converges to the accumulation point $P(0)$ of the set $A$, another subsequence converges to the isolated point $P'(2)$ of  $A.$
Note: 
1) Point $0$ and point $2$, are both accumulation points of the sequence $(x_n)_{ n \in \mathbb{N}}.$
2) Point $0$ is an accumulation point of $A$ , point $2$ is an isolated point of $A$.
