Formal proof that $\mathbb{R}^{2}\setminus (\mathbb{Q}\times \mathbb{Q}) \subset \mathbb{R}^{2}$ is connected. Cam anyone provide me the proof of:

that $\mathbb{R}^{2}\setminus (\mathbb{Q}\times \mathbb{Q}) \subset \mathbb{R}^{2}$ is connected.

 A: Theorem. $\ \mathbb{R}^2-A\ $ is connected for any set $A\subset\mathbb{R}^2$ of cardinality less than the continuum.
Proof. Consider any two points $u,v$ in $\mathbb{R}^2-A$. There is a foliation of continuum many paths from $u$ to $v$, which are disjoint except at $u$ and $v$. For example, one could consider all the various circle fragments containing $u$ and $v$. Since only fewer than continuum many of these paths contain points from $A$, it follows that almost all of them are contained in $\mathbb{R}^2-A$, which is therefore path-connected, and even arc-connected. QED
In particular, $\mathbb{R}^2-\mathbb{Q}\times\mathbb{Q}$ is connected. 
A: Suppose $(x,y),(a,b)\in \mathbb{R}^2\setminus\mathbb{Q}\times\mathbb{Q}$.  Then either $x$ or $y$ is irrational.  Suppose $x$ is irrational.  Then there's a path from $(x,y)$ to $(x,b)$ that remains in $\mathbb{R}^2\setminus\mathbb{Q}\times\mathbb{Q}$, namely, the second coordinate changes from $y$ to $b$ while $x$ stays put.  Now you have to get from $(x,b)$ to $(a,b)$ along a path.  If $b$ is irrational, you do it the same way except that it's the $x$-coordinate that changes, from $x$ to $a$.  If $b$ is rational, then you're going to need to put the "horizontal" path elsewhere than at $b$ and use two vertical paths instead of one.
A: In fact if $F\subset\mathbb{R}^{2}$ is such that $|F|<|\mathbb{R}|$ then $\mathbb{R}^{2}\setminus F$ is path connected. In this link there is a proof (connected of $\mathbb{R}^{2}\setminus F$). I think I am very formal in the demonstration.
