(Ref. page 9 of the following notes)
The problem is to prove that if $U \in \mathbb{R}^n$ is the feasible region of a linear program, then $x \in U$ is a basic feasible solution iff it's a vertex. Initial steps in the proof:
Suppose $x$ is a basic feasible solution and also that $x$ is a convex combination of $y$ and $z$, both in $U$. Let $C$ be a matrix whose rows are $n$ linearly independent active constraints at $x$, and $c$ be the vector of corresponding constraint values. Because $C$ has linearly independent rows, it has full rank and is invertible. Also $Cy \leq c, Cz \leq c$ and $Cx = c$ ...
$Cx = c$ is obviously correct by definition, but why are the other two inequalities valid: $Cy \leq c, Cz \leq c$ ? How is $C$ related in any way to $U$? If the feasible region is characterized by a polyhedron $\mathcal{P} = \{x \in \mathbb{R}^n\ |\ Ax \leq b\}$, then it's understandable to say that $Ay \leq b, Az \leq b$, since $y,z \in U$.
$C$ is solely defined as the matrix of linearly independent constraints active at $x$ and I don't see why it has anything to do with the feasible region.
Let $I := \lbrace i \vert A_{i \cdot} x = b_i \rbrace$ be the set of indices corresponding to the active constraints at $x$. Then $C$ is can be defined by $C := A_{I \cdot}$ and $c = b_I$ . Since $y$ is feasible we have $Cy = A_{I \cdot}y \leq b_I = c$ and similarly for $z$.
In your case the rows should be linearly independent and this is done by selecting an appropriate subset of $I$. In any case the constraints in $C$ are chosen from the constraints in $A$ and the above still holds.