I know that it's work if we remove two points $a,b$ of $S^2\subset \mathbb{R}^3$ he's still connected and for $S^1$ doesn't work, and start thinking if we can do for any $n\geq 3$ and for $n$ points, but i can't prove that. Also, if we can remove a countable points and he's still connected.
To see that $S^2-\{a,b\}$ is connected is see that he is homeomorphic to $S^1\times \mathbb{R}$, that is connected.
I don't know what tools you have at your disposal, but you should note that by stereographic projection, $S^n\setminus \{p\}\cong \mathbb{R}^n$. So, $S^n\setminus \{p_1,\ldots, p_n\}$ is homeomorphic to $\mathbb{R}^n$ minus a finite set of points. It is not very hard to see that $\mathbb{R}^n\setminus \{q_1,\ldots, q_m\}$ is path connected for $n\ge 2$ and any $m$. One way to do this is to explicitly construct a path between any pair of points in $\mathbb{R}^n\setminus \{q_1,\ldots, q_m\}$.