Suppose $U$ is an ordered infinite set, I want to construct subsets $A$ and $B$ such that:

(1) (Disjoint) $A \cap B = \phi $ 

(2) (Exhaustive) $A \cup B = U$ 

(3) (Alternate elements) $\forall x,y \in A, x<y, \exists z \in B $ such that $x < z<y$ and vice versa for B. 

(4) (Same cardinality) $|A|=|B|$ (i.e. There exists a bijection $h : A\rightarrow B$).

Can we construct $A$ and $B$ for $U = \mathbb{R}$ (set of real numbers)

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For $U = \mathbb{Z}$ (integers), we can take $A=\{2z\}, B = \{2z+1\}, z \in \mathbb{Z}$

For ${U} = \mathbb{Q}$ (set of rationals), we can define $A = \{(2m)/n\}, B = \{(2m+1)/n \}$ (where $m$ and $n$ have no common factor)

My main question is what if $U=\mathbb{R}$ (set of real numbers)? Can we construct $A$ and $B$ satisfying all 4 properties? And further on any arbitrary interval?