Let $x_k = (\sin k, \arctan k, k^3)$, $k \in \mathbb Z_+$. Prove (by the open ball definition) that $V = \mathbb R^3\setminus\{x_k: k \in \mathbb Z_+\}$ is open in $\mathbb R$ with the standard metric.
We consider $x \in V$ and define $B(x,r)$ such that $r = \text{min}\{|x-x_k|:k \in \mathbb Z_+\}$. Certainly no $x_i $ for any $i \in \mathbb Z_+$ can be in $B(x,r)$; if $x_i \in B(x,r)$, then $$|x - x_i| < \text{min}\{|x-x_k|:k \in \mathbb Z_+\} \leq |x - x_i|$$ which is a contradiction. Thus $B(x,r) \subset V$. Is my reasoning correct?
No, it is not correct. By that same argument, the complement of every countable subset of $\mathbb R^3$ would be open, and this is not true. For instance,$$\left\{\left(\frac1n,0,0\right)\,\middle|\,n\in\mathbb N\right\}^\complement$$ is not open in $\mathbb R^3$.
Your proof fails because that minimum that you mentioned doesn't exist in general. In your specific case it exists, but that's because $\lim_{k\to\infty}\lVert x_k\rVert=\infty$. Use this fact to complete the proof.