= Solution
Fix $z$ with $|z|<1-\varepsilon$, and let $\Pi(w)=w/\max(1,|w|)$ be radial projection to the closed disc. The <continuous map> $p\mapsto\Pi(p-G(p)+z)$ has a <fixed point> $p$ by the <Brouwer fixed-point theorem>. On the boundary, $|p-G(p)+z|<\varepsilon+|z|<1$, so its image under $\Pi$ is interior and cannot equal the boundary point $p$. Therefore $p\in B$. An interior <fixed point> of this projection must satisfy $p-G(p)+z=p$, because points outside the disc project to its boundary. Thus $G(p)=z$. \b[Every $z\in B_{1-\varepsilon}(0)$ belongs to $G(B)$.]
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