In the book The Geometry of Domains in Spaces by Krantz and Parks, the authors proved the weak $(1,1)$-type estimate of the maximal function $M_\mu f$, where $\mu$ is a Radon measure, using their version of the Besicovitch covering theorem.
Let $d$ be a positive integer. Then there exists a constant $C=C(d)$ such that for any finite collection of balls $\mathcal B = \{B_i\}_{i=1}^m$ in $\Bbb R^d$ with the property that no ball contains the center of any other ball, we can partition the family $\mathcal B$ into $$ \mathcal B = \mathcal B_1 \cup\mathcal B_2 \cup \dots \cup \mathcal B_C, $$ where each subfamily $\mathcal B_j$ consists of disjoint balls.
This version of the covering theorem seems pretty restrictive, especially the part about no ball contains the center of any other balls. Indeed, in theor proof of the weak $(1,1)$-type estimate, they relied on a certain claim that they did not prove.
Edit As Skeeve mentioned in the comment, this claim is not explicitly stated in the book but more of a paraphrasing of the part the authors left out in a proof.
Claim: Let $K\subset \Bbb R^d$ be a compact set such that each $x\in K$ is associated with a real number $r_x>0$. Then $K$ can be covered by a family of balls $$ \mathcal B = \{ B(x_i,r_i) : i=1,\dots,k\ \}, $$ where $r_i := r_{x_i}$, such that for any distinct $i,j \le k$, we have $$ x_i\notin B(x_j,r_j) \quad\text{and}\quad x_j\notin B(x_i,r_i). $$
I don't find this claim to be trivial at all. In fact, I tried many different methods but failed to prove it. Note that the mapping $x\mapsto r_x$ doesn't enjoy any nice property like continuity of any kind.
While the usual version of Besicovitch covering theorem circumvents this problem, I still would like to know how to prove the above claim (or a counter example if it is actually false).