I am reading the proof of Lévy–Steinitz theorem from ON THE POWER OF LINEAR DEPENDENCIES, which asserts that given a finite set $V$ of the unit ball $B$ of any norm(in $\mathbb{R}^d$) such that $\sum_{v\in V}v = 0$, there is a permutation of the elements of $V$, such that the norm of all the prefix sums are bounded by $d$.
In the course of the proof they consider the set $P$ of auxiliary functions $\beta:V_k\rightarrow[0,1]$ such that
\begin{align} &\sum_{v\in V_k}\beta(v)v = 0 \\ &\sum_{v\ V_k}\beta(v) = k-1-d \end{align}
Here $V_k\subset V$ of size $k$. $P$ is a convex polytope in the unit cube $[0,1]^k$. Let $\beta^\star(\cdot)$ be an extreme point of $P$. Now it is claimed that for some $v\in V_k$, $\beta^\star(v) = 0$.
Assume not. Now the auxiliary system has $d+1$ equations and $k$ variables. So at least $k-(d+1)$ of the iequalities $\beta^\star(v)\leq 1$ are satisfied as equalities. Then it is shown that this leads to a contradiction.
Now if $P$ is full dimensional, I understand this must be true, since an extreme point is characterized by intersection of $k$ hyperplanes (there can be more, but never less than this). However if $P$ is not full diemnsional, I fail to see how this is true. For example consider a square in the $xy$-plane where the ambient space is 3-dimensional.
So my question is that, shouldn't one also prove that $P$ is full dimensional (assuming I am not mistaken)?
Thanks