In a metric space a sequence with no converging subsequences is discrete (?) I've been trying to prove that given a metric space $X$ (not necessarily complete) and a sequence $(x_n)_n \subseteq X$ which contains no convergent subsequences, there exists an open neighborhood $V_n$ of $x_n$ for each $n \in \mathbb{N}$ such that these $V_n$'s are pairwise disjoint i.e.:
$$
\exists V_1 \in \mathcal{v}_{x_1}, \ldots, \exists V_n \in \mathcal{v}_{x_n}, \ldots:
\forall n_1 \neq n_2: V_{n_1} \cap V_{n_2} = \emptyset.
$$
So far I am stuck trying to prove it by contradiction. All I've managed to see is that it wouldn't be enough to consider balls of the same radius around each element of the sequence, as this wouldn't prove anything for the sequence $(1+1/1, 1+1/2, \ldots, n+\frac{1}{n}, n+\frac{1}{n+1}, \ldots)$ in $\mathbb{R}$, which doesn't have any convergent subsequences. So I'm trying to work with the general hypothesis by contradiction:
$$
\forall r_1, \ldots, r_n, \ldots: \exists n_1 \neq n_2:
B(x_{n_1}, r_{n_1}) \cap B(x_{n_2}, r_{n_2}) \neq \emptyset,
$$
but I am stuck.
Obviously, I could just pick for each $x_n$ a radius small enough $r$ so that $B(x_n, r)$ doesn't intersect some balls $B(x_1, r_1), \ldots B(x_{n-1}, r_{n-1}), B(x_{n+1}, r_{n+1}), \ldots$, because if that weren't posible, it would mean the sequence would have a converging subsequence to $x_n$. But the problem is, I can't guarantee that the balls obtained this way also don't intersect each other, not just $B(x_n, r)$.
I feel like there is something important I should be observing here, but I can't see it. Also, in case you happen to know that this result isn't true or even better, you happen to have a counterexample, it would be much appreciated. Thank you for reading.
 A: You're pretty close.  As you noted, you can find radii $r_n$ such that $B(x_n, r_n)$ does not contain any other $x_k \ne x_n$.  Now consider the balls $B(x_n, r_n/2)$.  By a triangle inequality argument, you should be able to show that these balls are pairwise disjoint.
A: For each $x_i$ in the sequence, consider its distances from the other points in the sequence. Define the numbers
$$d_i=\inf_{j\neq i}\{d(x_i,x_j)\}$$
If this infimum was 0, then we could find a subsequence converging to $x_i$, which is a contradiction. Therefore, we know that each $d_i$ must satisfy $d_i>0$. Consider the collection of balls
$$\mathcal{B}=\{B_{d_i/2}(x_i)\}$$
Can you show that these balls are disjoint?
A: If you assume the $x_n$ to be pairwise distinct, it can work. If $x_1 = x_2$, you are going to have trouble finding $V_1$ and $V_2$.
A: Assuming you can prove it for complete metric spaces, here's how to proceed for non-complete ones. Let $X$ be the given (non-complete) metric space and let $\overline{X}$ be its completion. Then $(x_n)$ is also a sequence in $\overline{X}$, and you can find open subsets $V_i$ of $\overline{X}$. With the desired propreties. Now the sets $V_i \cap X$ are open subsets of $X$ with the desired propreties.
