In a metric space a sequence with no converging subsequences is discrete (?)

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.

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?

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 $V_2$.