Every open subset $O$ of $\Bbb R^d,d \geq 1$, can be written as a countable union of almost disjoint closed cubes. After the usual construction as explained in press.princeton.edu/chapters/s8008.pdf (page 8) I did not understand the argument why the union is all of $O$. "We note that given $x \in O$ there exists a cube of side length $2^{-N}$ (obtained from successive bisections of the original grid) that contains $x$ and that is entirely contained in $O$" (page 8 line 4). How can this cube be rejected ? it is entirely contained in $O$. 
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 A: Let ${\cal S}_n$ be the set of all closed cubes of side length $2^{-n}$ obtained from successive bisections of the grid ${\mathbb Z}^d\subset{\mathbb R}^d$. The set ${\cal S}:=\bigcup_{n\geq0}{\cal S}_n$ is countable.
Let an nonempty open set $\Omega\subset{\mathbb R}^d$ be given. 
We shall construct  an increasing sequence $(B_n)_{n\geq0}$ of "buildings" $B_n\subset\Omega$. Each $B_n$ will be an almost disjoint union of cubes $C\in{\cal S}$ of sidelength $\geq2^{-n}$. Start with  $B_{-1}:=\emptyset$ and proceed recursively as follows:
Given $B_{n-1}$ for an $n\geq0$, let $B_n$ be the union of  $B_{n-1}$ with all cubes $C\in{\cal S}_n$ which are contained in $\Omega$, but not in $B_{n-1}$. These $C$ are almost disjoint with all cubes present in $B_{n-1}$. The limit set
$$\Omega':=\bigcup_{n\geq0} B_n$$
is then a countable union of almost disjoint cubes. I claim that $\Omega'=\Omega$.
Proof. The inclusion $\Omega'\subset\Omega$ is obvious.
For the converse, consider a point ${\bf x}=(x_1,\ldots, x_d)\in\Omega$. There is an $h>0$ such that the cube $K:=\prod_{i=1}^d [x_i-h,x_i+h]$ is contained in $\Omega$, and there is enough space to find an $n\geq0$ and a $C\in{\cal S}_n$,  such that
$${\bf x}\in C\subset K\subset\Omega\ .$$ When $C\not\subset B_{n-1}$ then $C \subset B_n$ by definition of $B_n$, so that in any case ${\bf x}\in C\subset B_n\subset\Omega'$. As ${\bf x}\in\Omega$ was arbitrary this proves $\Omega\subset\Omega'$.
