Proof of The Geometric Hahn-Banach Theorem on Brezis Page 7

Question

In the proof of the second geometric form of the Hahn-Banach Theorem on page 7 of Brezis's Functional Analysis, we established that

$$ f(x-y) \le f(rz) \quad \forall x\in A, y\in B, |z|=1.$$

From this, Brezis deduces that $$f(x-z)\le -r||f||.$$

1. Why can we do this? Is it because $f(rz)\le -r||f||$? But the statement is false. Rather, $-r||f||\le f(rz)\le r||f||$.

2. Can we fix this argument by requiring that $$ f(x-y) \le -2\epsilon \quad \forall x\in A, y\in B?$$

Given that $H=[g=\alpha]$ separates sets $C=A-B$ and $\mathcal{B}(0,r)$, we can write down a new linear functional $f=g-\alpha-2\epsilon$, so that $H=[f=-2\epsilon]$.

Thank you!