Seeing this post I wanted to expand the proof to half-open boxes, that is sets of the form $B = \prod_{i=1}^d [a_i,b_i)$, to get a completely disjoint union.
What I have done so far is adapting the proof which can be found page 7 of this paper :
Let $U \subseteq \mathbb{R}^d$ an open set. We would like to construct a collection $Q$ of half-open boxes such that $U = \bigcup_{B \in Q} B $.
We consider our space $\mathbb{R}^d$ with a grid formed by all half-open cubes with side length 1 and vertices at integer coordinates. To construct the collection $Q$ of cubes we either accept a cube $B$ if $B \subseteq U$, reject it if $B \subseteq U^c$ or partially accept it if $B$ intersects $U$ and $U^c$.
Then, we bisect all partially accepted cubes into $2^d$ cubes with side length $\frac{1}{2}$ and repeat this procedure indefinitely.
To check if $U$ is entirely covered by the collection $Q$, it is written :
we note that given x ∈ $U$ 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 $U$. Either this cube has been accepted, or it is contained in a cube that has been previously accepted.
I can't wrap my mind around this argument, I feel like I would be missing points since some sides of my cubes are open.
Any help towards an intuitive explanation would be welcome.
