Prove that $\mathbb R$ is not a nontrivial disjoint union of open subsets. I want to show that if we write $\mathbb R$ as a disjoint union of more than one subset of $\mathbb R$, then at least one of the subsets is not open in $\mathbb R$.
I'm sure I can prove this, but not very cleanly, by contradiction: if $\mathbb R$ were a nontrivial disjoint union of open subsets, then we could decompose each subset into its maximally-sized intervals of the form $(-\infty, a)$ or $(b, +\infty)$ or $(a, b)$ with $a < b$; then there exist two neighboring intervals in the overall decomposition, and the neighboring intervals do not contain their common boundary point, which would contradict the disjoint union covering $\mathbb R$.
Is there a better/cleaner/more elegant proof?

If this were possible, then it would be possible to write $\mathbb{R}$ as the union of two nonempty disjoint open subsets (since the union of any arbitrary collection of open sets is open).  This, however, is impossible.  Suppose $\mathbb{R} = V \cup W$, where $V$ and $W$ are nonempty open sets and $V \cap W = \emptyset$.
Now consider an element $x \in \overline{V} \setminus V$.  Can we have $x \in W$?
