Let $\mathbb{S}^k \subset \mathbb{R}^{k+1}$, $k \geq 1$, be the unit sphere.
Leopardi proved that, given a natural number $n \geq 2$, there exists a partition of $\mathbb{S}^k$ into pairwise disjoint sets $\{S_j\}_{j=1}^n\subset \mathbb{S}^k$, such that $$\sigma(S_j)=\frac{\sigma(\mathbb{S}^k)}{n}\quad \text{and} \quad \text{diam}(S_j)\leq \frac{c_1(k)}{n^{\frac{1}{k}}},$$ for all $j=1,\ldots,n$, where $c_1(k)>0$ is a positive constant only depending on the dimension $k$, diam$(\cdot)$ denotes the diameter and $\sigma(\cdot)$ the surface area. These partitions are called area regular partitions.
Given a point $x\in\mathbb{S}^k$ and $-1\leq t\leq1$, we define the spherical cap $$C=C(x,t)=\{y\in\mathbb{S}^k;\,\langle x,y\rangle\leq t\},$$ where $\langle\cdot,\cdot\rangle$ is the standard scalar product.
Fixed a spherical cap $C=C(x,t)$, it can be proved that the cardinality of the set of indices $\mathcal{J}(C) \subset \{1,\ldots,n\}$ such that, if $j\in\mathcal{J}(C)$ then $C\cap S_j \neq \emptyset$ and $(S^k\setminus C)\cap S_j \neq \emptyset$, is less or equal than a constant depending only on $k$, $c_2(k)$, times $n^{1-\frac{1}{k}}$, i.e., $$\text{card}(\mathcal{J}(C)) \leq c_2(k)n^{1-\frac{1}{k}}.$$
The proof I am following gives no detail: it only says that we have to use the properties of the partition of $\mathbb{S}^k$ (which it is quite obvious), but I do not know how to apply them. Any ideas or possible proofs?
For each such index $j$, the set $S_j$ intersect the boundary of the spherical cap, so it is contained in the $c_1(k)n^{-1/k}$-neighborhood of this boundary. The spherical area of this neighborhood is bounded above by $c(k) n^{-1/k}$ (maximized when it is a hemisphere), and since the $S_j$ are disjoint and all have the same spherical area, you get $$\textrm{card}(\mathcal{J}) \frac{\sigma(S^k)}{n} \le \frac{c(k)}{n^{1/k}},$$ which implies what you want to show.