Decomposing the real line into everywhere isomorphic suborders

We show that if $\mathbb{R} = A \cup B$ is a partition of $\mathbb{R}$ into two suborders $A$ and $B$, then there is an open interval $I$ such that $A \cap I$ is not order-isomorphic to $B \cap I$. The proof depends on the completeness of $\mathbb{R}$, and we show in contrast that there is a partition of the irrationals $\mathbb{R} \setminus \mathbb{Q} = A \cup B$ such that $A \cap I$ is isomorphic to $B \cap I$ for every open interval $I$. We do not know if there is a partition of $\mathbb{R}$ into three suborders that are isomorphic in every open interval.