Suppose we have a compact orientable surface $S$. Let $A \subset S$ be a discrete subset. Let the the set of accumulation points of $A$ be denoted by $B$, is $B \setminus A$ discrete?
(A point $p$ is called an accumulation point of $A$ point if every open subset of $S$ containing $p$ intersects $A$)
As I pointed out in the comment, you can find a similar counter-example in which $B\setminus A$ is not discrete.
Considert the discrete set $A=\{(\frac{1}{m},\frac{1}{n})\mid m,n\in\mathbb{Z}_+\}$ in the real plane $\mathbb{R}^2$. It has infinitely many limit points so that its closure $B=A\cup\{(\frac{1}{m},0)\mid m\in\mathbb{Z}_+\}\cup\{(0,\frac{1}{n})\mid n\in\mathbb{Z}_+\}\cup\{(0,0)\}$.
Notice that $B$ has one non-discrete point, $\{(0,0)\}$. Now it remains to embed $A\subset\mathbb{R}^2$ into the $2$-sphere $S^2$ to satisfy the condition of your question, but it is trivial (e.g., think of stereo projection).