Let $A$ and $B$ be $n \times n$ Hermitian matrices and denote by $F_A$ and $F_B$ the distribution functions related to the spectral measures $L_A$ and $L_B$ of $A$ and $B$, respectively. Setting $k = \mathrm{rank}(A-B)$, prove the following rank inequality $$\|F_A - F_B\|_\infty \leq \frac{k}{n}.$$ This problem is taken from Exercise 6.3 of this book where the authors want us to deliver a proof based on Cauchy's interlacing theorem. I am wondering whether there is an alternate (perhaps simpler) route to this "linear-algebra" result?

Some background: Let $H$ be a $n \times n$ Hermitian matrix with real eigenvalues $\lambda_1(A) \geq \lambda_2(A) \geq \cdots \geq \lambda_n(A)$, the spectral measure of $A$, denoted by $L_A$, is just the counting measure on the eigenvalues of $A$, i.e., $L_A = \frac{1}{n}\sum_{i=1}^n \delta_{\lambda_i(A)}$. Given any pair of probability measures (or probability distribution functions) $\mu, \nu$ on $\mathbb{R}$, the distance $$\|\mu - \nu\|_\infty := \sup\limits_{x\in \mathbb R} \left|\mu\left((-\infty,x)\right) - \nu\left((-\infty,x)\right) \right|$$ is also known as the Kolmogorov distance between $\mu$ and $\nu$ (denoted by $\mathrm{d}_K(\mu,\nu)$)