Prove that the set $ \{x_n: n \in \mathbb{N} \} $ has no accumulation points. Suppose $(x_n) $ is a sequence in a metric space $ (X, d) $ such that for all
$ n \neq m, \quad d (x_n, x_m) \geq r $ for some $ r> 0 $ fixed. Prove that the set $ \{x_n: n \in \mathbb{N} \} $
has no accumulation points.
My try: 
Suppose that for all $\quad r > 0 \quad B_r(x_m) - \{x_m\}\cap A \neq \emptyset$,
then $\{x_m\}\subset A$, which is a contradiction because $n \neq m$
I'm not sure if my process is correct. Any suggestions would be great! 
 A: Suppose that $x$ is an accumulation point and consider the ball $B_{r/2}(x)$. It should contain infinitely many $x_n$'s, but the distance between any two elements of $B_{r/2}(x)$ is smaller than $r$, by the triangle inequality.
A: Denote by $X'$ the set of accumulation points of $X$ and $B(a,r)$ by the open ball centered at $a$ with radius $r>0$. The following propositions will help you to conclude your problem. 

If $a\in X'$, then every ball centered in $a$ contains an infinity many points of $X$. 

Indeed, given $B(a,r)$, since $a\in X'$ then exist $x_1\neq a$ such that $x_1\in X\cap B(a,r)$. Let $r_1=d(a,x_1)$. Exist $x_2\neq a$ such that $x_2\in X\cap B(a,r_1)$. Let $r_2=d(a,x_2)$. We have $0<r_2<r_1$. Exist $x_3\neq a, x_3\in X\cap B(a,r_2)$. Let $r_3=d(a,x_3)$. Continuing this fashion we can define infinity many points $x_1,x_2,x_3,\dots$ in $X\cap B(a,r)$.

If $a\in X'$, then $a$ is the limit of a sequence of points in $X$.

For every $n\in\Bbb N$ the open ball $B(a,1/n)$ contains infinitely many points of $X$. Then we can choose the points $x_1,x_2,\dots,x_n,\dots$ successively in such way that $x_n\in X\cap B(a,1/n)$ and $x_n$ is different from $x_1,\dots,x_{n-1}$ chosen before. Observe that $m\neq n\Rightarrow x_n\neq x_m$ and by definition $d(x_n,a)<1/n$ then $\lim x_n=a$.
Now set $X=\{x_n;n\in\Bbb N\}$ and for $m\neq n, d(x_n,x_m)\geq r$, for some $r>0$. If $a\in X'$, then exist a sequence $(y_n)$ in $X$ such that $\lim y_n=a$, that is, given $r>\epsilon>0$ exist $n_0\in\Bbb N$ such that $n>n_0\Rightarrow d(y_n,a)<\epsilon/2$. Then, for $n,m>n_0$ we get
$$d(y_n,y_m)\leq d(y_n,a)+d(a,y_m)<\dfrac{\epsilon}{2}+\dfrac{\epsilon}{2}<r.$$
But $(y_n)$ is a sequence in $X$ then $d(y_n,y_m)\geq r$. And we have a contradiction, so $X$ has no accumulation points.
