Every open interval in $\mathbb{R}$ can't be written as the union of disjoint closed sets.
This implies that every open sets in $\mathbb{R}$ can't be written as the union of disjoint closed sets.
I wonder if this is also true for $\mathbb{R}^n$ but I have trouble take the proof of $n=1$ to $n\ge2$.
Indeed, this cannot be done: no (nonempty) open set in $\mathbb{R}^n$ can be written as a countable union of disjoint closed sets. This is a consequence of the Baire category theorem; GEdgar's answer to this question gives the details.
Let me give more details.
Suppose for a contradiction that $U\subseteq\mathbb{R}^n$ is open and $U=\bigcup_{n\in\mathbb{N}} C_n$, with $C_n$ closed and $C_i\cap C_j=\emptyset$ for $i\not=j$.
Let $B_n$ be the boundary of $C_n$, and $B=\bigcup B_n$. (Formally, the boundary of a closed set is the set of all points in the closed set which are not in the interior of the closed set; note that a closed set can be equal to its own boundary, as in the case of the Cantor set.) Then you can show that:
$B$ (viewed as a subspace of $\mathbb{R}^n$) is a complete metric space.
Each $B_n$ is a nowhere dense subset of $B$.
But this is a contradiction, since by the Baire Category Theorem no complete metric space is a countable union of nowhere dense sets.
