Two cubes in $\mathbf{R}^n$ are almost disjoint if the intersection of their interior is empty. It is covered in Tao's "Introduction to Measure Theory" book that
Lemma 1.2.11: Every open set $E \subseteq \mathbf{R}^d$ can be covered by countably many almost disjoint dyadic cubes.
I shall first present the proofs from the book and post the question after the proof:
Question:
I am not certain how $\mathcal{Q}^*$ is guaranteed to be nonempty. In other words, what guarantees us that a maximal set from $\mathcal{Q}$ even exists at the first place? My idea is to use the Zorn's lemma and treat $\mathcal{Q}$ as a partially ordered set with set inclusion ordering. However, I am stuck at showing $\mathcal{Q}$ is inductive. That is, given a totally ordered subset of $\mathcal{Q}$, how do we start to define an upper bound for this set?
Take $Q$ to be a cube of side length $2^{-n}$ in $\mathcal{Q}$. By what was said about parent cubes, we can inductively define $Q_i$ for integers $i$, $0 \leq i \leq n$ by the following rule:
$Q_n = Q$. Having defined $Q_i$, $Q_{i-1}$ is the unique "parent" cube of side length $2^{-i+1}$ containing $Q_i$.
We can define $S = \{m \in \mathbb{Z}^+ : 0 \leq m \leq n, Q_m \subset E\}$. $S$ has a minimum element, say $N$. In terms of set inclusion, $Q_N$ is maximal (if it were contained in a larger dyadic cube, itself contained in $E$, then the unique parent cube of $Q_N$ would also be contained in $E$, which would contradict the fact that $N$ is the minimum of $S$). So $Q_N \in \mathcal{Q}^*$, and $\mathcal{Q}^*$ is nonempty.