Any open subset of $\mathbb R$ is an uniquely countable union of disjoint open intervals I know how to prove that any open subset of $\mathbb R$ is a countable union of disjoint open intervals. See this question. 
I would like to prove this union is unique. Is there a straightforward way to prove this?
 A: It essentially follows from the fact that open intervals are connected spaces.
Let $S$ be open open subset of $\mathbb R$. Assume that $\{U_i\}_{i\in I}$ and $\{V_j\}_{j\in J}$ are two such representations of $S$ as the union of open intervals.
Fix $i\in I$, and note:
$$U_i = \bigcup_{j\in J} (U_i\cap V_j)$$
Then the $U_i\cap V_j$ are disjoint open subsets of $U_i$. But $U_i$ is connected. So only one of $U_i\cap V_j$ can be non-empty, and thus $U_i\cap V_j=U_i$ for some $V_j$ - that is $U_i\subseteq V_j$.
Then prove similarly that for each $j\in J$ there must be exactly one $i$ so that $V_j\subseteq U_i$ and $V_j\cap U_{i'}=\emptyset$ if $i'\neq i$.
Use these results to prove that the two partitions are the same.
We are essentially reducing it to the case where $S$ is an open interval - connectedness then shows that $S$ can't be represented as the non-trivial union of disjoint open sets.
A: For $x\in S$ let $V(x)$ be the set of all open intervals $J$ such that $x\in J\subset S.$ Let $U(x)=\cup V(x).$ Each $U(x)$ is an open interval containing $x.$ And $\cup_{x\in S}U(x)=S.$ 
Let  $\{U_i:i\in I\}$ be a pair-wise disjoint family of open intervals with $\cup_{i\in I}U_i=S$  and such that $U_i \ne U_{i'}$ for distinct $i,i'\in I.$ 
For $x\in S$ there  exists a unique $j(x)\in I$ such that $x\in U_{j(x)}.$ We have $U_{j(x)}\in V(x)$ so $U_{j(x)}\subset \cup V(x)=U(x).$ 
Suppose $U_{j(x)}\ne U(x)$. Because $U_{j(x)}\subset U(x)$ we may let $y=\sup U_{j(x)}$ if $\sup U_{j(x)}\in U(x)\setminus U_{j(x)},$ or, otherwise, let $y=\inf U_{j(x)}.$ In either case we have $y\in U(x)\setminus U_{j(x)}.$ But $y\in S$ because $y\in U(x)\subset S,$ so there exists $r>0$ and some $i\in I\setminus \{j(x)\}$  such that $(-r+y,r+y)\subset U_{i}.$ From the definition of $y$ it should be  clear that this implies $U_i\cap U_{j(x)}\ne \phi,$ which is absurd because $i\ne j(x).$ 
Therefore, by contradiction, we have $U_{j(x)}=U(x).$  
Therefore $\{U_i:i\in I\}=\{U_{j(x)}:x\in S\}=\{V(x):x\in S\}.$
