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?
 A: 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 $x \in W$?
A: (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. 
A: The statement is equivalent to proving that $\mathbb{R}$ is not the union of two disjoint nonempty open sets. Indeed, if $\mathbb{R}$ is the disjoint union of nonempty open sets (more than one), then taking one of them and the union of the others, you obtain $\mathbb{R}$ as the disjoint union of two nonempty open sets.
Suppose $\mathbb{R}=A\cup B$, with $A$ and $B$ disjoint, open and nonempty. Since the complements of open sets are closed, $A$ and $B$ are also closed.
Let $a\in A$ and $b\in B$; without loss of generality, $a<b$.
Consider $S=\{x\in\mathbb{R}:a\ge x, [a,x]\subseteq A\}$. The set $S$ is upper bounded (by $b$) and not empty (because $a\in S$). Therefore $s=\sup S$ exists. Since $A$ is closed and $S\subseteq A$, we have that $s\in A$, because the supremum of a set belongs to its closure.
I claim that $s\in S$. Indeed, if $[a,s]\not\subseteq A$, then there is $t\in [a,s]\cap B$ and $a<t\le s$. Since $B$ is open, there is $\varepsilon>0$ such that $(t-\varepsilon,t+\varepsilon)\subseteq B$. Note that $t-\varepsilon>a$, because $a\in A$. On the other hand, since $s=\sup S$, there is $x\in S$ with $t-\varepsilon<x\le s$, a contradiction because $[a,x]\subseteq A$.
Since $A$ is open, there exists $\delta>0$ such that $(s-\delta,s+\delta)\subseteq A$. But then $[a,s+\delta/2]\subseteq A$, contradicting the fact that $s=\sup S$.

This is the proof that $\mathbb{R}$ is connected. A straightforward modification shows that every interval in $\mathbb{R}$ is connected.
