$S^2$ with countably many points removed is path-connected 
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?
 A: Here is a solution which doesn't use any machinery from algebraic topology.

Let $X'$ denote the countable set of points removed from $S^2$. Let $x \in X'$ and let $X = X'\setminus\{x\}$. By stereographic projection from $x$, $S^2\setminus X'$ is homeomorphic to $\mathbb{R}^2\setminus Y$ where $Y$ is the image of $X$ under the stereographic projection. As path-connectedness is invariant under homeomorphism, it is enough to show that $\mathbb{R}^2\setminus Y$ is path-connected.
Let $p_1, p_2$ be distinct points in $\mathbb{R}^2\setminus Y$. I claim there is a line $L_1$ which passes through $p_1$ such that $L_1\cap Y = \emptyset$; note, $L_1\cap Y = \emptyset$ is equivalent to $L_1 \subseteq \mathbb{R}^2\setminus Y$. If not, there would be at least one point of $Y$ on each line through $p_1$; this is a contradiction as there are uncountably many such lines, but only countably many such points. This argument actually shows there are uncountably many lines which pass through $p_1$ which do not meet $Y$.
If $p_2 \in L_1$, then we see that there is a continuous path (in $\mathbb{R}^2\setminus Y$) from $p_1$ to $p_2$ obtained by travelling along $L_1$. 
If $p_2 \not\in L_1$, by the above reasoning, there is a line $L_2$ in $\mathbb{R}^2$ through $p_2$ with $L_2 \cap Y = \emptyset$. As there are uncountably many such lines, we can choose $L_2$ such that it is not parallel to $L_1$ (there is only one line through $p_2$ which is parallel to $L_1$). Any two non-parallel lines in $\mathbb{R}^2$ meet in a unique point, so $L_1\cap L_2 = \{q\}$. Now we see there is a path from $p_1$ to $p_2$ (in $\mathbb{R}^2\setminus Y$) given by travelling from $p_1$ to $q$ along $L_1$, then travelling from $q$ to $p_2$ along $L_2$.
As any two points $p_1, p_2 \in \mathbb{R}^2\setminus Y$ can be joined by a continuous path in $\mathbb{R}^2\setminus Y$, we see that $\mathbb{R}^2\setminus Y$ is path-connected. Therefore, $S^2\setminus X'$ is path-connected.

The above argument can be slightly altered to prove $S^n\setminus X'$ is path-connected for $n \geq 2$. The only difference is that lines need to be replaced by hyperplanes (note that lines are hyperplanes in $\mathbb{R}^2$). The key point is that you will always be able to find hyperplanes $H_1$, $H_2$ through $p_1$, $p_2$ respectively which do not meet $Y$, and we can choose them to be non-parallel so that they must intersect.
A: 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 $X$.
