Definition: A convex polyhedral cone is a set $$ \sigma = \{r_1 v_1 + \dots + r_s v_s \in V \mid r_i \ge 0\}$$ A face $\tau$ of $\sigma$ is the intersection of $\sigma$ with any supporting hyperplane $$ \tau = \sigma \cap u^\perp = \{v \in \sigma \mid \langle u,v \rangle = 0\}$$ for some $u \in \sigma^\vee$
A facet is a face of codimension one.
I have a question about a specific part of the proof of the proposition "any proper face is contained in some facet."
It suffices to show that if $\tau = \sigma \cap u^\perp$ has codimension greater than one, it is contained in a larger face. WLOG, $\sigma$ spans $V$. Let $W$ be the linear span of $\tau$. The images $\overline{v_i}$ in $V/W$ of the generators of $\sigma$ are contained in a half-space determined by $u$.
By moving this half-space in the sphere of half-spaces in $V/W$, one can find one that contains these vectors $\overline{v_i}$, but with at least one such nonzero vector in the boundary hyperplane. In other words, there is a $u_0$ in $\sigma^\vee$ so that $u_0 ^\perp$ contains $\tau$ and at least one of the vectors $v_i$ not in $W$; this means that $\sigma \cap u_0^\perp$ is a larger face.
My confusion is in the second paragraph. How do we actually "move" half-spaces and why does $V/W$ determine a sphere of half-spaces? Secondly, what is the "boundary hyperplane"?
I'm not really looking for concrete arguments, but rather a general understanding of what the idea of the proof is.
The boundary hyperplane of a half space $\{u \geq 0\}$ defined as the topological boundary of this set - i.e. thee set $\{u=0\}$.(To prove this just take any sequence $x_n \neq x$ converging to $x$ in $\{u=0\}$ and negate the $x_n$ in the sequence for which $\lambda(x_n)<0$ to get $x_n \to x$ with $x_n \notin \{u \geq 0\}$.)
To answer your second question: Any vector space $V$ of $n$ dimensions, determines a sphere of half spaces: Here is what he means by the term - any half space in $V$ determines a linear functional in $V^*$ up to a scalar multiple. From the identification of oriented grassmanians $\check G_{n-1}(V) \cong \check G_1(V)$, identify these half spaces with oriented hyperplanes and identify this with $G_1(V)$ which is just the the spherization of $V^*$.
Here is an 'all at once' proof that I hope will clear the confusion: Take a basis $\tau_1...\tau_k$ of $W$, with $\{\tau_i\}_{i \leq k} \in \tau$, and continue it to a basis of $V$, using $\sigma_{k+1},...\sigma_{n}$, where $\sigma_k \in \sigma$. Consider $V/(W \oplus span\{\sigma_{k+1},...\sigma_{n-1} \})$ and the image $\overline{\sigma}$ under the quotient. $\check{\overline{\sigma}} \neq 0$ since otherwise $\sigma$ would be a vector space, which it isn't, again since $\overline{\sigma}$ is strongly convex. Thus there is a (unique up to scalar multiplication) $ \lambda \in \check {\overline{\sigma}}$. Viewing $\lambda$ as $ \in V^*$ via $ \left(V/(W \oplus span\{\sigma_{k+1},...\sigma_{n-1} \})\right)^* \cong \operatorname{Ann}(W \oplus span\{\sigma_{k+1},...\sigma_{n-1} \}) \subset V^*$, we have $\lambda>0$ on the basis vector $\sigma_{n}$ and equal to $0$ on $W \oplus span\{\sigma_{k+1},...\sigma_{n-1} \}$. Thus $span\{\{\lambda=0\}\cap \sigma \}$ is of codimension $1$ in $V$, so $\{\lambda=0\} \cap \sigma$ is a facet containing $\tau$.