Is $[0,1]^2 \setminus \{(a,b)\}$ connected? I am pretty sure that this set is in fact connected but I am struggling to see how to prove it, it is simple to see that $[0,1] \setminus \{x\}$ is disconnected but I can't see how to relate techniques used in that example to this one.
 A: Consider that, between any two points $(x_1,y_1)$ and $(x_2,y_2)$, there are at least two distinct paths, sharing no points other than the endpoints. For instance, take the line segment $(x_1,y_1)$ and $(x_2,y_2)$ and, for any $(x_3,y_3)$ not colinear with $(x_1,y_1)$ and $(x_2,y_2)$, draw the line segments $(x_1,y_1)$ to $(x_3,y_3)$ then to $(x_2,y_2)$.
A: I will show path connectedness for a more general problem.
Begin with the unit square. Remove a countable number of points. Let $p$ and $q$ be any two points left in the unit square. Consider the uncountable set of all lines going through $p$ restricted to the unit square. Remove all of the lines that go through some of the countable points that were removed. We are still left with an uncountable number of lines. Denote these lines by $\mathscr{L}_p$ and similarly $\mathscr{L}_q$.
Prove by contradiction that the points on the lines in $\mathscr{L}_p$ are dense in $[0,1]^2$. That means, in any neighborhood of $[0,1]^2$ there is a point that lies on one of the lines in $\mathscr{L}_p$. The easiest way to do this is to use convexity and state that if this weren't true then more than a countable number of lines would have been removed. Naturally, the same is true for $\mathscr{L}_q$. 
Now, construct a path that takes $p$ arbitrarily close to $(0,0)$ and then arbitrarily close to $(1,1)$. Construct a path that takes $q$ arbitrarily close to $(1,0)$ and then arbitrarily close to $(0,1)$. As long as you go close enough to the vertices of the square, the paths are going to intersect and you can claim path connectedness.
I wrote out more details than necessary but I think it is still pretty straightforward. It is an interesting exercise to adapt this proof to certain suitable non-convex sets and to prove it without relying on coordinates.
