TITLE: Arcwise connected part of $\mathbb R^2$
QUESTION [19 upvotes]: Here's a question that I share:
 Show that if $D$ is a countable subset of $\mathbb R^2$ (provided with its usual topology) then $X=\mathbb R^2 \backslash D $ is arcwise connected.

REPLY [2 votes]: Ok I will take a stab at clarifying Brian M. Scott's Answer instead of writing a fully different answer.
Let $p,q\in \mathbb{R}-D$. Then Define $A:=\{\text{ Lines Through }p\}$, $B:=\{ \{\text{ Lines through }q\}$.
One could write the definitions of these sets more formally if desired, but the meaning is clear. I claim the following:
$$
\exists l\in A, \text{ such that } l\cap D=\emptyset. 
$$
Similarly
$$
\exists n\in B, \text{ such that } l\cap D=\emptyset. 
$$
Lets us understand why this should be clear. Now define a mapping $m:\mathbb{R} \to A$  by  $m(x):=$ the line in $A$ with slope $x$. This mapping is clearly $1-1$ and onto (almost see below*), as any line through a point $p$ is determined by its slope, and any slope uniquely determines a line through $p$. You could formalize this more by writing any line through $p$ in point slope form, but I leave that to the reader.
A similar argument follows for $B$, thus $||A||=||B||=||\mathbb{R}||$.
Now let us show that there are lines in $A$ and $B$ which do not contain points in $D$. define a map $\phi:D\to A\times B$, by $\phi(d):=\{(l,m)\in A\times B| d=l\cap m\}$. That is each point $d\in D$ gets mapped to the pair of lines in $A$ and $B$ which meet at $d$. This mapping is $1-1$, as any two points determine a unique line.
By assumption $D$ is countable, therefor $\phi$ cannot be onto thus there exists a pair of lines, $(l^*,m^*)$, which do not contain any points in $D$. This follows as $A\times B- \phi(D)\neq \emptyset$.
we have a piecewise linear path $p \to l^*\cap m^* \to q$. QED.
The Crux of the argument, here is where we assert that the mapping $\phi$ cannot be onto, given the assumption that $D$ was countable.
One small note, the map $m$ is not quite onto, as the vertical line through $p$ has undefined slope. However, this doesn't disturb the argument as the main idea is that $A$ and $B$ are both uncountable. We might also be able to circumvent this by taking a mapping from the extended reals.