It has been shown that every open set in ℝ is the union of disjoint open intervals. Is this true for $ℝ^n$ in general? It has been shown that every open set in $\mathbb R$ is the union of disjoint open intervals. Is this true for $\mathbb R^n$ in general?
It seems to me that it is not, but I cannot think of an example.
 A: It's unclear what you mean by "interval" in a higher-dimensional context. Reasonable candidates include boxes ( = Cartesian products of open intervals) and open balls. However, here's a general result showing that almost any such claim will fail even in $\mathbb{R}^2$:
Let $\mathcal{O}$ be some family of open sets (e.g., all open boxes), and let $U$ be a connected open set which is not an element of $\mathcal{O}$ (e.g., $U$ is an open ball). Then $U$ cannot be written as a disjoint union of elements of $\mathcal{O}$. Why? Well, suppose it could - as $U=\bigsqcup A_i$. Fix some nonempty $A_i$ in particular. Since $U\not\in\mathcal{O}$, we have $A_i\subsetneq U$; so (why? this is where connectedness of $U$ gets used!) we can find a point $x$ on the boundary of $A_i$, which is in $U$. Then any open set at all containing $x$ has to intersect $A_i$. But $x\in A_j$ for some $j$, since $x\in U$! So $A_i\cap A_j\not=\emptyset$, contradiction.
EDIT: Or, as Bungo points out below, $A_i$ and $\sqcup_{j\not=i} A_j$ partition $U$ into two disjoint open sets, contradicting the assumption that $U$ was connected. 
The key difference between the two-(or higher-)dimensional case and the one-dimensional case? In one dimension, the connected open sets are easy to classify - they're exactly the open intervals! In two dimensions, though, there are lots of connected open sets. Really, the theorem that works for all dimensions is

Any open set can be written as a disjoint union of connected open sets;

it's just that for $n=1$ this can be phrased in a particularly snappy way.
