Breaking down a maximal $r$-separated subset of $\mathbb R^d$ into a finite disjoint union of $s$-separated subsets Let $A$ be a maximal $r$-separated subset of $\mathbb R^d$, meaning the distance of any two points in $A$ is at least $r>0$.
claim: For any $s>0$, we can always write $A$ as a finite disjoint union of $s$-separated subsets $A_1,\dots, A_p$ such that $p$ only depends on the ratio $s/r$ and the dimension $d$. (Thanks to the comments below. To be more precise, I want the minimal possible value of $p$ among all decompositions to be bounded by a function that only depends on $s/r$ and $d$. )
When $s\le r$, we can simply take $p=1$ and $A=A_1$.
When $s>r$, my idea is that we first take a maximal $s$-separated subset $A_1$ of $A$ and then take a maximal $s$-separated subset $A_2$ of $A-A_1$ and so on. But I don't see why this has to stop in finite steps and the number of steps only depends on $s/r$ and $d$. For high dimensions this is not easy to picture.
In the original problem I am working on $s/r$ is an integer, but I guess that is not necessary. Please let me know if it matters. This might be related to some famous theory I don't know about.
 A: I give below an argument to show that $A$ is indeed a $\textit{finite}$ union of $s$-separated subsets. But you claim needs to be more precise.
Assume $s>r$. Let $N$ be the maximal number of disjoint balls of radius $r$ that can be put inside a ball of radius $s+r$ (such a number does exist). $N$ depends on $s/r$ and $d$ only.
You can construct as you did, thanks to Zorn's Lemma, a family $(A_i)_{i \in \mathbb{N}}$ of subsets of $A$ such that $A_{i+1}$ is a non empty maximal $s$-separated subset of $A \setminus (A_1 \cup \cdots \cup A_{i})$, or if $A=A_1 \cup \cdots \cup A_{i}$, take $A_{i+1}=\emptyset$.
Let's show that there is only finitely many non empty $A_i$'s. Assume $A_{N+2} \neq \emptyset$ and take $x \in A_{N+2}$. Since $A$ is $r$-separated in $\mathbb{R}^d$, the ball $B(x,s)$ meets at most $N$ $A_i$'s. So there exists an integer $1 \leqslant n \leqslant N+1$ such that $B(x,s) \cap A_n = \emptyset$. But then $A_n \cup \{ x \}$ is in $A \setminus (A_1 \cup \cdots \cup A_{n-1})$, is $s$-separated and $A_n \neq A_n \cup \{ x \}$ , which contradicts the maximality of $A_n$ in $A \setminus (A_1 \cup \cdots \cup A_{n-1})$. This shows that $A_{N+2}=\emptyset$ and implies by construction $A=A_1 \cup \cdots \cup A_{N+1}$ ($A_{N+1}$ may be empty but for the finiteness it's okay).
