 Solution

Fix $z$ with $|z|<1-\epsilon$, 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, $|p-G(p)+z|<\epsilon +|z|<1$, 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 $G(p)=z$. Every $z\in B_{1-\epsilon }(0)$ belongs to $G(B)$.