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First, as I mentioned in the comments, proposition b pretty clearly needs a separability assumption in order for the sub cover $\mathcal G$ to be countable. I can prove the proposition, provided we interpret the discreteness criterion of the set of radii in a strong sense. That is, in the case where the radii are bounded, I need the second condition to imply that $\{r_B\mid B\in\mathcal{F}\}$ is well ordered by $\geq$, i.e., every subset of radii has a maximal element.

First, as I mentioned in the comments, proposition b pretty clearly needs a separability assumption in order for the sub cover $\mathcal G$ to be countable. I can prove the proposition, provided we interpret the discreteness criterion of the set of radii in a strong sense. That is, in the case where the radii are bounded, I need the second condition to imply that $\{r_B\mid B\in\mathcal{F}\}$ is well ordered by $\geq$, i.e., every nonempty subset of radii has a maximal element.

Finally, we formally adopt the convention that a (closed) ball $B$ is just a pair $(x_0,r)$ consisting of a center $x_0 \in X$ and a radius $r>0$, even though we will often abuse notation and identify $B$ with the point set $\{x\in X\mid d(x,x_0)\leq r\})$. Thus we will write things like “$x\in B$“, but behind the scenes, it is important that we have a designated center and radius for every ball, since in a metric space a ball’s center and/or radius need not be uniquely determined by its point set. We denote by $c(B)$ and $r_B$ the respective center and radius of the ball $B$.

Finally, we formally adopt the convention that a (closed) ball $B$ is just a pair $(x_0,r)$ consisting of a center $x_0 \in X$ and a radius $r>0$, even though we will often abuse notation and identify $B$ with the point set $\{x\in X\mid d(x,x_0)\leq r\}$. Thus we will write things like “$x\in B$“, but behind the scenes, it is important that we have a designated center and radius for every ball, since in a metric space a ball’s center and/or radius need not be uniquely determined by its point set. We denote by $c(B)$ and $r_B$ the respective center and radius of the ball $B$.

With that out of the way, here's a proof:

To see this, observe that the first condition implies $$\chi_A\leq \Sigma_{B\in\mathcal G}\chi_B\text{,}$$ while on the other hand, the second condition implies that for any $x\in\bigcup \mathcal G$, $\{B\in\mathcal G \mid x\in B\}$ is a Besicovitch family, and so has cardinality at most $K$, proving that $$\Sigma_{B\in\mathcal G}\chi_B\leq K\text{.}$$

$$\mathcal G_B = \begin{cases} \bigcup_{B'\prec B} \mathcal G_{B'} & \text{if } c(B)\in \bigcup_{B'\prec B} \mathcal G_{B'}\text{,}\\ \bigcup_{B'\prec B} \mathcal G_{B'} \cup \{B\} & \text{otherwise.} \end{cases} $$

With that out of the way, here's a proof, which invokes separability only to establish countability of $\mathcal G$:

To see this, observe that the first condition implies $$\chi_A\leq \sum_{B\in\mathcal G}\chi_B\text{,}$$ while on the other hand, the second condition implies that for any $x\in\bigcup \mathcal G$, $\{B\in\mathcal G \mid x\in B\}$ is a Besicovitch family, and so has cardinality at most $K$, proving that $$\sum_{B\in\mathcal G}\chi_B\leq K\text{.}$$

\begin{equation} \mathcal G_B = \begin{cases} \displaystyle\bigcup_{B'\prec B} \mathcal G_{B'} & \text{if } c(B)\in \displaystyle\bigcup\left(\displaystyle\bigcup_{B'\prec B} \mathcal G_{B'}\right)\text{,}\\ \left(\displaystyle\bigcup_{B'\prec B} \mathcal G_{B'}\right) \cup \{B\} & \text{otherwise.} \end{cases} \end{equation}