I'm wondering if the following strengthening of the Besicovitch covering theorem holds: Suppose $A\subset\mathbb R^n$ is a bounded subset and suppose $x\mapsto r_x$ is a function $A\to(0,\infty)$. Is it possible to choose constants $0<\lambda<1$ and $N\in\mathbb N$ (both depending only on the dimension $n$) so that there exists a countable subset $C\subset A$ such that the following two statements hold?

(1) the collection of balls $\{B(x,\lambda r_x)\}_{x\in C}$ covers $A$ and,

(2) we have the pointwise inequality $\sum_{x\in C} \chi_{B(x,r_x)}\le N$ (where $\chi_E$ denotes the indicator function of a set $E$). In words, no point of $\mathbb R^n$ is contained in more than $N$ of the balls $B(x,r_x)$ (with $x\in C$).

If we replaced $\lambda$ by $1$, this statement would be true as a consequence of the usual Besicovitch covering theorem. I'm somewhat stuck trying to modify that proof to get this stronger statement (I would just like $\lambda$ to be strictly smaller than $1$, it can be as close to $1$ as we like).

**Is this stronger version with some $0<\lambda<1$ true? If not, is there an easy counterexample? If it is true, a proof (or sketch/hint/reference) would be appreciated!**

If it helps to answer the question, I don't mind adding the additional assumption that $A$ is actually a compact set (i.e., its closed in addition to being bounded) and $x\mapsto r_x$ is a continuous (or even Lipschitz) function.