I’m struggling to understand the last part of this proof where it says $\pi(x)=\pi(x-z)$ proves the claim. To prove the claim I suppose this must imply that $||x|| \leqslant ||x-z||$ so that $||x||<1$ also and thus $x$ belongs to the open unit ball in $E$.
I can’t see how this implies this.
In case this is not standard, I have in my notes $\pi : E \rightarrow E/E_0$ , $x \mapsto x+E_0$ is the canonical projection where $E_0$ is a closed subspace of the normed space $E$ and the quotient norm is defined to be $||x +E_0||:= inf\{||x-z||:z\in E_0\}$
It's enough to find some $y\in E$ with $\|y\|\lt 1$ such that $\pi(y)=x+E_0$.( In other words, the preimage of $x+E_0$ intersects the open unit ball in $E$ non-trivially.) The element $x-z$ is such a choice.