Just out of curiosity, is it possible to partition $\mathbb{R}$ into two dense sets of equal cardinality?
I was thinking something like this: Let $S$ be the basis of $\mathbb{R}$ over $\mathbb{Q}$. Then $S$ is of equal cardinality as $\mathbb{R}$. Let $A$ be an uncountable subset of $S$ such that $S\setminus A$ is also uncountable. Such a set exists, for example if $f$ is a bijection from $S$ to $\mathbb{R}$, put $a$ in $A$ if and only if $f(a)>0$. Now let $T$ be the subset of $\mathbb{R}$ such that $t\in T$ if and only if $tq\in A$ for some rational number $q$. Then $T$ and $\mathbb{R}\setminus T$ gives the desired partition.
But, as far as I know, the existence of a basis of $\mathbb{R}$ over $\mathbb{Q}$ depends on the Axiom of Choice. Is it possible to prove the above statement without assuming the Axiom of Choice?