to say $x\not\in {\mathcal{C}}_r(u)$.

3.  We have the estimate $\sum r_i^{n-2} < C(n,\lambda)^d$.

Ignoring the construction of the balls $B_{r_i}(x_i)$ for a moment, let us remark that we are done if we can always find such a collection. Indeed, in this case we can then consider each ball $B_{r_i}(x_i)$ independently. If $r_i=r$ we leave the ball alone, otherwise by rescaling $r_i\to 1$ and translating $x_i\to 0$ we now have a ball $B_1(0)$ such that $r^{d-1}_0\geq 1$, and hence we can find a $d-1$-covering as above for the new ball. In particular, this means we can cover $B_{r_i}(x_i)$ by a collection of balls $B_{r_{ij}}(x_{ij})$ which satisfy the above conditions for $d-2$. Summing over all $i$ and $j$ gives us a collection of balls $\{B_{r_{ij}}(x_{ij})\}\subseteq B_1(0)$ of our original ball which satisfy (1) and (2) above and for which $$\begin{aligned}
\sum r_{ij}^{n-2}\leq C(n,\lambda)^{d}\cdot C(n,\lambda)^{d-1}\, .\end{aligned}$$ Continuing this $d$ times we arrive at a collection of balls $B_r(y_j)$ which satisfy (2) and for which $\sum r^{n-2} \leq C^{\frac{1}{2}d(d-1)}\leq C^{\Lambda^2}$, which finishes the proof.

Hence, we are left with understanding the construction of the balls $\{B_{r_i}(x_i)\}\subseteq B_1(0)$ satisfying (1),(2),(3) above under the assumption that $r_0^{d}\geq 1$. Roughly, the construction proceeds as follows. For every 