TITLE: $S^2$ with countably many points removed is path-connected
QUESTION [5 upvotes]: Prove that after removing a countably infinite number of points from $S^2$, it remains path-connected.

This was a question that arose in the algebraic topology course I have this term. I thought geometrically about that. I think it makes sense if we don't take it hard, but how should I show it in logical and rigid way?

REPLY [4 votes]: Here is a slightly more general proposition: For any connected manifold $M$ of dimension $\geq 2$, and a countable subset $X$, $M - X$ is path connected.
Let $p,q$ be two points in $M - X$. Let $\gamma$ be a path between them in $M$. By compactness, definition of a manifold, and a cardinality argument, $\gamma$ can be decomposed as a sequence of paths $\gamma_1,...,\gamma_n$ with each $\gamma_k$ contained in a coordinate patch $U_k$ homeomorphic to $\mathbb{R}^n$ by a map $\psi_k$ and such that the endpoints of $\gamma_k$ are not in $X$ (this last bit the cardinality argument, which for full rigor actually requires some tedious invocations of continuity of $\gamma$ and connectedness of the unit interval). 
Now, consider each $\psi_k \circ \gamma_k$ living in $\mathbb{R}^n$ and after a linear transformation we may assume the endpoints are $(0,0,...,0)$ and $(1, 0, 0, ... ,0)$. Then define paths $$\eta_{k,t}(s) = \psi_k^{-1}((s, t\sin(\pi s),0,0,...,0)$$
(The intuition is just to create an uncountable family of paths disjoint except at the endpoints), and observe that for each $k$ there is some $t_k$ such that $\eta_{k,t_k}$ doesn't hit $X$ by cardinality concerns. Then the path composition of $\eta_{k,t_k}$ is a path $p$ to$q$ not hitting any point of $X$.