The question
Let $I$ be the unit $n$-cube in $\mathbb{R}^n$, defined by the inequalities $\{0\leq x_i\leq 1 \, |\, i=1,2,...,n\}$. Define $\partial_d I$ to be its $d$-dimensional skeleton, or, in other words, the union of the $d$-dimensional faces of $I$.
Denote by $\pi_V : \mathbb{R}^n \to V$ the orthogonal projection onto $V$, where $V$ is a $d$-dimensional linear subspace of $\mathbb{R}^n$.
Is it true, then that $$ \pi_V(\partial_d I) = \pi_V(I) $$ for any $V$?
A sketch of a proof.
Let $p \in \pi_V(I)$. Then $\pi^{-1}(p)$ is an $n-d$-dimensional plane intersecting $I$. This intersection is a closed convex polytope, which we denote by $P$, and it's contained in $\pi^{-1}(\{p\})$. Let $v$ be a vertex of $P$. By vertex we mean a point in $P$ which is \textit{not} contained in the interior of any $k$-dimensional face of $P$, with $k \geq 1$. By translating and rotating, we may assume that $v=0$. If $0\in \partial_d I$, we are done. Suppose not. Then $0$ belongs to the interior of some $\ell$-dimensional face of $I$, where $\ell \geq k+1$. Denote this face by $F$. We can find a ball $B$ centered at $0$, so that $B \cap F$ contains $\ell$ linearly indepedependent vectors $\{v_F\}$. Similarly, because $P$ is an $n-k$ polytope, we can find $n-k$ lineraly independent vectors $\{v_P\}$ in $B\cap P$. By convexity of $P$ and the fact that $0$ is a vertex of $p$, $F \cap P = \{0\}$. In particular, any linear combination of the vertex found in $P\cap B$ cannot lie in $F$. We deduce that $\{v_F\} \cup \{v_P\}$ is a family of at least $n-k + k+1 = n+1$ linearly independent vectors in $\mathbb{R}^n$. This is impossible, and hence $0$ must lie in a $k$ dimensional face of $I$ with $k\leq d$.
The issue
All the statements in the sketch of the proof seem obviously true to me, except the final statement that $\{v_P\} \cup \{v_F\}$ is a linearly independent family of vectors. I still think that it's true, but I can't think of any quick proof. Any comment is welcome!
Also, if you have a reference where the fact I'm trying to prove is shown, that would be great!
Thank you!