Quite often there is a common trick in harmonic analysis like the following:
Assume that $0<S(x)<\infty$ for all $x\in\mathbb{R}^{n}$. Consider the set \begin{align*} \Omega_{k}&=\{x\in\mathbb{R}^{n}: S(x)>2^{k}\},\\ \mathcal{D}_{k}&=\left\{\text{dyadic cube}~\mathcal{Q}\subseteq\mathbb{R}^{n}:|\mathcal{Q}\cap\Omega_{k}|>\dfrac{1}{2}|\mathcal{Q}|~\text{and}~|\mathcal{Q}\cap\Omega_{k+1}|\leq\dfrac{1}{2}|\mathcal{Q}|\right\},\quad k\in\mathbb{Z}. \end{align*} The author claimed without further reasoning that for each dyadic cube $\mathcal{Q}\in\mathcal{D}_{k}$, there is a maximal dyadic cube $\mathcal{R}\in\mathcal{D}_{k}$ such that $\mathcal{Q}\subseteq\mathcal{R}$.
I fail to see why such a statement is valid. He also mentioned that this is a stopping time argument, perhaps the labeling does not matter that much, what bothers me is the maximality argument.
There are several facts regarding the properties of $\mathcal{D}_{k}$. The union $\displaystyle\bigcup_{k\in\mathbb{Z}}\mathcal{D}_{k}$ of all $\mathcal{D}_{k}$ contains all the dyadic cubes in $\mathbb{R}^{n}$, this is not too hard to check. Besides that, if $\mathcal{Q}\in\mathcal{D}_{k}$ for some $k$, then $\mathcal{Q}\in\mathcal{D}_{j}$ for all $j\ne k$, this is easy to check. So the crucial question is that, why the following cannot hold? \begin{align*} \mathcal{Q}_{1}\subseteq\mathcal{Q}_{2}\subseteq\cdots\mathcal{Q}_{i}\subseteq\cdots, \end{align*} each $\mathcal{Q}_{i}\in\mathcal{D}_{k}$ and the chain does not terminate. We know that for any dyadic cubes $\mathcal{Q},\mathcal{R}$, either they are disjoint or they contain each other: $\mathcal{Q}\subseteq\mathcal{R}$ or $\mathcal{R}\subseteq\mathcal{Q}$. So if the chain does not terminate, then there is no such maximal dyadic cube.