[proofplan] The construction is a stopping-time on the dyadic grid: for each $x \in \Omega$, select the largest dyadic cube containing $x$ whose diameter is at most $\delta(x)/4$, where $\delta(x) = \operatorname{dist}(x, \Omega^c) > 0$. Maximality ensures disjointness via the nesting property of dyadic cubes; the lower bound on $\operatorname{dist}(Q_j, \Omega^c)$ in (ii) comes from the stopping condition; the upper bound in (ii) comes from the fact that the dyadic parent failed to be selected. The size-comparability (iii) for cubes whose doubles intersect follows by chaining the upper bound on $\operatorname{dist}(Q_j, \Omega^c)$ for the smaller cube through the shared point $z \in Q_j^* \cap Q_k^*$ to the lower bound on $\operatorname{dist}(Q_k, \Omega^c)$ for the larger one, yielding $\ell_k \le 11\,\ell_j$. The dyadic constraint $\ell_k/\ell_j \in 2^{\mathbb{Z}_{\ge 0}}$ then forces $\ell_k/\ell_j \le 8$, giving the ratio bound $1/8 \le \ell(Q_j)/\ell(Q_k) \le 8$. Throughout, "diam" is the Euclidean diameter $\sqrt{n}\,\ell(Q)$, where $\ell(Q)$ is the side length. [/proofplan]

[step:Define the stopping condition and select the maximal admissible dyadic cubes] For $x \in \mathbb{R}^n$, let $\delta(x) := \operatorname{dist}(x, \Omega^c) = \inf\{|x - y| : y \in \Omega^c\}$. The function $\delta: \mathbb{R}^n \to [0, \infty)$ is $1$-Lipschitz, and since $\Omega \subsetneq \mathbb{R}^n$ is open, $\delta(x) > 0$ if and only if $x \in \Omega$. For a dyadic cube $Q \in \mathcal{D}$, recall $\ell(Q) = 2^{-k}$ if $Q$ is of generation $k$, and $\operatorname{diam}(Q) = \sqrt{n}\,\ell(Q)$. Call $Q \in \mathcal{D}$ **admissible** if \begin{align*} Q \subseteq \Omega \quad \text{and} \quad \operatorname{diam}(Q) \le \operatorname{dist}(Q, \Omega^c). \end{align*} Call an admissible $Q$ **selected** if no strict dyadic ancestor $Q' \supsetneq Q$ is admissible. Define \begin{align*} \mathcal{S} := \{Q \in \mathcal{D} : Q \text{ is selected}\}, \qquad \{Q_j\}_{j \in J} := \mathcal{S}, \end{align*} enumerating $\mathcal{S}$ by some countable index set $J$ (countable because $\mathcal{D}$ is countable). [/step]

[step:Prove the selected cubes are pairwise disjoint and partition $\Omega$] By the **nesting property of dyadic cubes**, any two dyadic cubes are either disjoint or one contains the other. Suppose $Q_i, Q_j \in \mathcal{S}$ with $Q_i \cap Q_j \ne \varnothing$ and, without loss of generality, $Q_i \subseteq Q_j$. If $Q_i \subsetneq Q_j$, then $Q_j$ is a strict dyadic ancestor of $Q_i$ that is admissible (since $Q_j \in \mathcal{S}$ implies admissibility), contradicting the maximality clause for $Q_i \in \mathcal{S}$. Hence $Q_i = Q_j$. For coverage: take any $x \in \Omega$, so $\delta(x) > 0$. There is a smallest generation $k_0 \in \mathbb{Z}$ such that the generation-$k_0$ dyadic cube $Q_{k_0}(x)$ containing $x$ is admissible (this exists because for $k$ large, $Q_k(x)$ has $\operatorname{diam}(Q_k(x)) = \sqrt{n}\,2^{-k}$ small relative to $\delta(x)$, while for $k$ very negative, $Q_k(x)$ is large and fails $Q_k(x) \subseteq \Omega$). By construction $Q_{k_0}(x) \in \mathcal{S}$, and $x \in Q_{k_0}(x)$. Hence $\Omega \subseteq \bigcup_{j} Q_j$. Conversely each selected cube lies in $\Omega$ by admissibility, so $\bigcup_j Q_j \subseteq \Omega$. Combined with disjointness, $\Omega = \bigsqcup_j Q_j$, proving (i). [/step]

[step:Verify the lower bound $\operatorname{diam}(Q_j) \le \operatorname{dist}(Q_j, \Omega^c)$ from admissibility] By construction every selected cube $Q_j$ is admissible, so the lower bound \begin{align*} \operatorname{diam}(Q_j) \le \operatorname{dist}(Q_j, \Omega^c) \end{align*} is the very definition of admissibility. This is the lower bound in (ii). [/step]

[step:Derive the upper bound $\operatorname{dist}(Q_j, \Omega^c) \le 4\operatorname{diam}(Q_j)$ via the parent failing admissibility] Let $Q_j \in \mathcal{S}$ and let $\tilde Q_j$ be its dyadic parent: the unique dyadic cube with $Q_j \subsetneq \tilde Q_j$ and $\ell(\tilde Q_j) = 2\ell(Q_j)$, hence $\operatorname{diam}(\tilde Q_j) = 2\operatorname{diam}(Q_j)$. By maximality of $Q_j$, the parent $\tilde Q_j$ is not admissible. Two reasons can cause non-admissibility: either $\tilde Q_j \not\subseteq \Omega$, or $\operatorname{diam}(\tilde Q_j) > \operatorname{dist}(\tilde Q_j, \Omega^c)$. **Case 1: $\tilde Q_j \not\subseteq \Omega$.** Then there is a point $z \in \tilde Q_j$ with $z \in \Omega^c$. Pick any $x \in Q_j \subseteq \tilde Q_j$. The Euclidean distance $|x - z| \le \operatorname{diam}(\tilde Q_j) = 2\operatorname{diam}(Q_j)$. Hence $\delta(x) \le |x - z| \le 2\operatorname{diam}(Q_j)$. Since $\operatorname{dist}(Q_j, \Omega^c) = \inf_{x \in Q_j} \delta(x) \le \delta(x) \le 2\operatorname{diam}(Q_j)$, \begin{align*} \operatorname{dist}(Q_j, \Omega^c) \le 2\operatorname{diam}(Q_j) \le 4\operatorname{diam}(Q_j). \end{align*} **Case 2: $\tilde Q_j \subseteq \Omega$ but $\operatorname{diam}(\tilde Q_j) > \operatorname{dist}(\tilde Q_j, \Omega^c)$.** Then $\operatorname{dist}(\tilde Q_j, \Omega^c) < \operatorname{diam}(\tilde Q_j) = 2\operatorname{diam}(Q_j)$. Since $Q_j \subseteq \tilde Q_j$, for any $x \in Q_j$ and any $y \in \Omega^c$ realising (up to $\varepsilon$) $\operatorname{dist}(\tilde Q_j, \Omega^c)$ from some $\tilde x \in \tilde Q_j$, \begin{align*} |x - y| \le |x - \tilde x| + |\tilde x - y| \le \operatorname{diam}(\tilde Q_j) + \operatorname{dist}(\tilde Q_j, \Omega^c) + \varepsilon < 2\operatorname{diam}(Q_j) + 2\operatorname{diam}(Q_j) + \varepsilon = 4\operatorname{diam}(Q_j) + \varepsilon. \end{align*} Letting $\varepsilon \to 0$, $\operatorname{dist}(Q_j, \Omega^c) \le 4\operatorname{diam}(Q_j)$. Combining both cases, $\operatorname{dist}(Q_j, \Omega^c) \le 4\operatorname{diam}(Q_j)$, the upper bound in (ii). [/step]

[step:Bound the distance from $Q_j$ to $Q_k$ via the closest-point projection of $z$] Let $Q_j, Q_k \in \mathcal{S}$ with $Q_j^* \cap Q_k^* \ne \varnothing$, where \begin{align*} Q^* := \{x : \operatorname{dist}_\infty(x, \text{centre}(Q)) \le \ell(Q)\} \end{align*} is the cube concentric with $Q$ and side length $2\ell(Q)$. Without loss of generality, $\ell(Q_j) \le \ell(Q_k)$ (the other case is symmetric); we will show $\ell(Q_k) \le 8\ell(Q_j)$. Write $\ell_j := \ell(Q_j)$, $\ell_k := \ell(Q_k)$, so $\operatorname{diam}(Q_i) = \sqrt{n}\,\ell_i$. Pick a point $z \in Q_j^* \cap Q_k^*$; by definition $\|z - c_j\|_\infty \le \ell_j$ and $\|z - c_k\|_\infty \le \ell_k$, where $c_i$ is the centre of $Q_i$. **Sharp distance bound from $z$ to each cube.** Since $\|z - c_j\|_\infty \le \ell_j$ and $Q_j = \{x : \|x - c_j\|_\infty \le \ell_j/2\}$, the $\ell^\infty$-distance from $z$ to $Q_j$ is at most $\ell_j - \ell_j/2 = \ell_j/2$ (attained on a face of $Q_j$). The Euclidean distance is at most $\sqrt{n}$ times the $\ell^\infty$-distance, so the closest point $\pi_j(z)$ of $Q_j$ to $z$ satisfies \begin{align*} |z - \pi_j(z)| \le \frac{\sqrt{n}\,\ell_j}{2}. \end{align*} Symmetrically, the closest point $\pi_k(z)$ of $Q_k$ to $z$ satisfies $|z - \pi_k(z)| \le \sqrt{n}\,\ell_k/2$. [/step]

[step:Derive $\ell_k \le 11\,\ell_j$ from the upper bound on $\operatorname{dist}(Q_j, \Omega^c)$] By the upper bound from step 4 applied to $Q_j$, for any $\varepsilon > 0$ there exist $w_j \in Q_j$ and $y \in \Omega^c$ with \begin{align*} |w_j - y| \le \operatorname{dist}(Q_j, \Omega^c) + \varepsilon \le 4\,\operatorname{diam}(Q_j) + \varepsilon = 4\sqrt{n}\,\ell_j + \varepsilon. \end{align*} We now bound $\operatorname{dist}(Q_k, \Omega^c)$ by chaining $\pi_k(z) \to z \to \pi_j(z) \to w_j \to y$: \begin{align*} \operatorname{dist}(Q_k, \Omega^c) &\le |\pi_k(z) - y| \le |\pi_k(z) - z| + |z - \pi_j(z)| + |\pi_j(z) - w_j| + |w_j - y| \\ &\le \frac{\sqrt{n}\,\ell_k}{2} + \frac{\sqrt{n}\,\ell_j}{2} + \operatorname{diam}(Q_j) + 4\sqrt{n}\,\ell_j + \varepsilon \\ &= \frac{\sqrt{n}\,\ell_k}{2} + \frac{\sqrt{n}\,\ell_j}{2} + \sqrt{n}\,\ell_j + 4\sqrt{n}\,\ell_j + \varepsilon \\ &= \sqrt{n}\,\Bigl(\frac{1}{2}\,\ell_k + \frac{11}{2}\,\ell_j\Bigr) + \varepsilon. \end{align*} Here we used $|\pi_j(z) - w_j| \le \operatorname{diam}(Q_j) = \sqrt{n}\,\ell_j$ since both $\pi_j(z)$ and $w_j$ lie in $Q_j$. Letting $\varepsilon \to 0$, \begin{align*} \operatorname{dist}(Q_k, \Omega^c) \le \sqrt{n}\,\Bigl(\frac{11}{2}\,\ell_j + \frac{1}{2}\,\ell_k\Bigr). \end{align*} Combining with the admissibility lower bound from step 3, $\sqrt{n}\,\ell_k = \operatorname{diam}(Q_k) \le \operatorname{dist}(Q_k, \Omega^c)$: \begin{align*} \sqrt{n}\,\ell_k \le \sqrt{n}\,\Bigl(\frac{11}{2}\,\ell_j + \frac{1}{2}\,\ell_k\Bigr) \iff \frac{\ell_k}{2} \le \frac{11\,\ell_j}{2} \iff \ell_k \le 11\,\ell_j. \end{align*} [/step]

[step:Upgrade to $\ell_k/\ell_j \le 8$ via the dyadic constraint and conclude property (iii)] Both $\ell_j$ and $\ell_k$ are powers of $2$ (members of the dyadic grid $\mathcal{D}$), and $\ell_j \le \ell_k$, so the ratio $\ell_k/\ell_j$ is a non-negative integer power of $2$: \begin{align*} \frac{\ell_k}{\ell_j} \in \{2^m : m \in \mathbb{Z}_{\ge 0}\} = \{1, 2, 4, 8, 16, \dots\}. \end{align*} Combined with $\ell_k/\ell_j \le 11$ (and $2^4 = 16 > 11$), \begin{align*} \frac{\ell_k}{\ell_j} \in \{1, 2, 4, 8\}, \end{align*} so in particular $\ell_k \le 8\,\ell_j$. **Symmetric inequality.** Reversing the roles of $j$ and $k$ in the argument of the previous step (using $\operatorname{dist}(Q_k, \Omega^c) \le 4\operatorname{diam}(Q_k)$ from step 4 and $\operatorname{dist}(Q_j, \Omega^c) \ge \operatorname{diam}(Q_j)$ from step 3) gives $\ell_j \le 11\,\ell_k$, and the dyadic constraint upgrades this to $\ell_j \le 8\,\ell_k$. Combined with $\ell_k \le 8\,\ell_j$, \begin{align*} \frac{1}{8} \le \frac{\ell(Q_j)}{\ell(Q_k)} \le 8, \end{align*} which is property (iii). [/step]

[step:Combine the four properties to conclude the theorem] The disjointness of $\{Q_j\}$ established in step 2, together with $\Omega = \bigsqcup_j Q_j$ from the same step, gives (i). The two-sided estimate $\operatorname{diam}(Q_j) \le \operatorname{dist}(Q_j, \Omega^c) \le 4\operatorname{diam}(Q_j)$ from steps 3 and 4 gives (ii). The size-comparability $\frac{1}{8} \le \ell(Q_j)/\ell(Q_k) \le 8$ for cubes with intersecting doubles, from steps 5–7, gives (iii). The collection $\{Q_j\}_{j \in J}$ is countable as a subset of the countable dyadic grid $\mathcal{D}$. This completes the proof. [/step]