Is $\Bbb Q\times(\Bbb R\setminus\Bbb Q)$ connected? Reading the definition for a set to be connected, I found that my intuition breaks down and I wonder whether the following set is connected?$
\def\Q{\Bbb Q}
\def\R{\Bbb R}
\def\c{^{\mathsf C}}
\def\-{\!\setminus\!}
\def\X#1#2{\mathop{\LARGE \times}_{#1}^{#2}}
\def\ue{\mathrm{u.\!e.}}
$ The complement is denoted as $\cdot\c$

Let $M\subset\R$ be countable and dense in $\R$. Is $X=M\times M\c$ connected?

I'd guess that the answer is independent of $M$, i.e. it does not matter whether it is the Rational numbers, the Algebraic numbers, or some real number-field etc.?
In the case that it's not connected:


*

*Can connectedness be achieved by adding more copies of $M\c$? Like is $X_n(\Q)$ connected for $$ X_n(M) = M\times\X 1 n M\c $$

*Can connectedness be achieved by using a set $M$ such that $M$ and $M\c$ are "uncountable everywhere"? Let "$M$ is u.e. in $\R^n$" be defined as:$$
M\subset\R^n \text{ is }\ue \quad\iff\quad (S\subset\R^n, S \text{ open } \implies |S\cap M| > \aleph_0)$$

Note: There is an other question if $\Q^n \cup (\R\-\Q)^n$ is connected, but I think that does not help here because it's union, not Cartesian product?

Suppose that $p,q \in \mathbb{R}$ are such that $p \notin M$ and $q \notin M^c$. Such numbers clearly exist. Then, $(M \times M^c) \cap (\{p\} \times \mathbb{R}) = (M \times M^c) \cap (\mathbb{R} \times \{q\}) = \emptyset$. Hence you can write $M \times M^c$ as the product of four (and hence also two) disjoint opens. 
On your first bullet point: if a set is connected, then all of its projections must be too. The answer to your second bullet point is also no, because I can just take the product of the nonalgebraic numbers by themselves. That's also not connected by the argument above.  
