Let $f:B_2(0)\to\mathbb C$ be holomorphic with $f(1)=1$ and $f(-1)=-1$. Show that there is a $z\in B_2(0)$ and a $\varepsilon>0$ s.t. $f(z)=1-\varepsilon$.
Is it reasonable to say that since $|f|$ has a local maximum at $z=1$ the function $f$ must be constant on all of $B_2(0)$ by the maximum principle and therefore we get that for $z=0$ and $\varepsilon=1$ it holds that $f(0)=1-1=0$ which is obviously constant.
If this is not enough on its own I was considering to show that $z=1$ has an open neighborhood in $f(B_2(0))$ however I am not sure how to argue that this must be true. Do you have hints for this?
I'm not really sure how you get the idea that $f(1)$ is a local maximum. You have that $f(-1)=-1$ and $f(1)=1$ so clearly $f$ is not constant. By the open mapping theroem (you can apply it since $f$ is not constant) $f(B_2(0))$ is open and $1 \in f(B_2(0))$ therefore there is a $\varepsilon$-ball around 1 with $B_{\varepsilon}(1)\subset f(B_2(0))$ But this is allready enough to show the claim.