Show that any open set $U\subseteq\mathbb{R}^n$ is a disjoint union of countably many intervals. 

As the title says, I have to show that any open set $U\subseteq\mathbb{R}^n$ is a disjoint union of countable many intervals.


Hello, my idea is the following: 
Consider any $x=(x_1,...,x_n)\in U\subseteq\mathbb{R}^n, U$ open. Then there exists a $\varepsilon>0$ so that $B(\varepsilon,x)\subseteq U$ because $U$ is open. Since $x$ is in the open interval
$$
(a,b)_x:=\left\{(x_1,...,x_n)\in\mathbb{R}^n|a_i<x_i<b_i, i=1,...,n\right\}
$$
with
$$
a_i:=x_i-\varepsilon,\quad b_i:=x_i+\varepsilon.
$$
Because of $\overline{\mathbb{Q}}=\mathbb{R}$, there exist $c_i,d_i\in\mathbb{Q}, i=1,...,n$ with
$$
a_i<c_i<x_i,\quad x_i<d_i<b_i.
$$
Define
$$
[c,d]_x:=\left\{(x_1,...,x_n)\in\mathbb{R}^n|c_i\leq x_i\leq d_i, i=1,...,n\right\},
$$
then $x\in [c,d]_x\subseteq U$.
Now take $x'\neq x$. If $x'$ is not in $[c,d]_x$, one can again construct a closed interval with rational endpoints, i.e., $[e,f]_{x'}$. There exists a $\varepsilon'>0$ with $B(\varepsilon',x')\subseteq U$. When minimizing $\varepsilon'$ to $\varepsilon''$ in such a way that 
$$
[c,d]_x\cap B(\varepsilon'',x')=\emptyset,
$$
$[c,d]_x$ and $[e,f]_{x'}$ are disjoint. If $x'$ is in $[c,d]_X$, choose $[e,f]_{x'}:=[c,d]_x$.
The desired countability is fulfilled because of the countability of $\mathbb{Q}$: There are countable many closed intervals with rational endpoints as constructed above.

I would like to know if my idea to proof is correct or nonsense.
With regards!
 A: It seems the following.
Your proof has a small technical gap (to provide $(a,b)_x\subset B(\varepsilon,x)$ you should take 
$a_i:=x_i-\varepsilon/\sqrt{n}$,$b_i:=x_i+\varepsilon/\sqrt{n}$ instead of 
$a_i:=x_i-\varepsilon$, $b_i:=x_i+\varepsilon$) and a principal idea error: you may not cover all points of the uncountable set $U$ by your countable inductive construction.
I see the general situation as follows. 


*

*As ABC already wrote in comments, you can cover each non-empty open subset $U$  of $\mathbb R^n$ by (countably many) disjoint open “intervals” only if $n=1$. 

*By Sierpiński Theorem (see Nuno’s answer), if
a continuum $X$ (that is a compact connected Hausdorff space) is a countable union $\bigcup_{i=1}^{\infty} X_i$ of its disjoint closed subsets then at most one of the sets $X_i$ is non-empty. As a corollary, you can cover no non-empty open subset $U$ of $\mathbb R^n$ by countably many disjoint closed “intervals”. 

*Any non-empty open subset of $\mathbb R^n$ is a disjoint union of countably many “halfclosed” cubes of the form  $2^{1-k}(x+[0;1)^n)$, where $k\in\mathbb N$ and $x\in\mathbb Z^n$. I used this construction to prove that each non-empty open subset of the space $\mathbb R^n$ is a disjoint union of $r$ homeomorphic parts if $r\ge 2^{n+1}-1$ or $n=2$ and $r\ge 4$ (see p. 2 of English draft version of my Ukrainian paper “Partitions of subsets of $\mathbb R^n$ onto similar sets” (Nauk. visn. Cherniv. univ., 269. Mathematics – Chernivtsi, Ruta, 2005 – P.88–93)).
A: Let $x \in U$ and let $U_x$ be the connected component of $U$ that contains $x$ that is - all $ a \in U$ that there is a continuous curve $\gamma:[0,1] \rightarrow U$  so 
$$\gamma(0)=x,\ \gamma(1)=a$ and $\gamma([0,1]) \subset U$$  
Since $U$ is open, $ \  \exists \  r_x > 0 $ so that an open ball $B(x,r_x)$ is contained by $U$. 
Open balls are connected sets so $x \in B(x,r_x) \subset U_x$ implies that $x$ is an interior point of $U_x$.
 Hence each connected component of $U$ can be represented by an open ball contained by it, and because these components are pairwise disjoints - so are the balls that represents them.
Each ball in this collection of representatives contains a rational point (all coordinates are rational numbers) hence each connected component can be represented by a unique n-length sequence of rational numbers thus at most countable many items in this collection.
