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?

(1) All intervals in $\mathbb{R}$ are connected. 
$$\mathbb{R} = \bigcup_{n \in \mathbb{Z}} (-n,n)$$
(2) Consider the diffeomorphism;
$$f: (-a,a) \to \mathbb{R}; \ x \mapsto \frac{ax}{\sqrt{a^2-x^2}}$$
Since all intervals of $\mathbb{R}$ are connected then $(-a,a)$ is connected. Moreover, since $f$ is continuous then $f(-a,a) = \mathbb{R}$ is connected. 
