The following is an excerpt from a paper by Kerman and Sawyer:
(The underlying space is $\mathbb{R}^{n}$).
Each cube $Q$ is covered by at most $2^{n}$ dyadic cubes $\{I_{j}\}_{1\leq j\leq 2^{n}}$ with $2^{-n}|Q|\leq|I_{j}|\leq|Q|$.
I am justifying this statement.
Let $i\in\mathbb{Z}$ be such that $2^{i}\leq\ell(Q)<2^{i+1}$. Denote by $S=\{I~{\rm dyadic}:\ell(I)=2^{i},I\cap Q\ne\emptyset\}$. The dyadic cubes $I$ in $S$ cover $Q$ and $|I|\leq|Q|$. Now we have \begin{align*} |Q|=\sum_{I\in S}|I\cap Q|. \end{align*} I am trying to argue that the number of elements of $S\leq 2^{n}$ but fail to do so.
I hope the following is correct.
Fix an $I_{j}\in S$. Consider any $I_{k}\in S$ in the first quadrant. Denote by $I_{k}'$ the parental dyadic cube that containing $I_{k}$. Then \begin{align*} \sum_{\substack{I\in S\\ I~\text{in the first quadrant}}}|I\cap Q|\leq|I_{k}'|=2^{n}|I_{k}|=2^{n}|I_{j}|. \end{align*} There are $2^{n}$ quadrants, decompose the sum $\sum_{I\in S}|I\cap Q|$ into quadrants sums, then we have the bound for $4^{n}$.