A matrix $A$ is $K$-irreducible if and only if no eigenvector lies on $\partial K$.

Question

I'm studying the properties of nonnegative matrices, and I encountered a theorem for which I can not understand its proof. The theorem can be found in "Nonnegative matrices in the Mathematical Sciences, Abraham P., Robert J.P.", ch 1, theorem 3.15.

I present what I believe is sufficient to understand the proof.

---

Definition Let $K \subset \mathbb{R}^n$. $K$ is a cone if $K=K^G$, where $K^G = \{\sum\limits_{k=1}^nc_iv_i \ |\ c_i \in \mathbb{R}_{\ge 0}, v_i \in K \}$, i.e. $K$ consists of all finite nonnegative linear combinations of itself.

Definition A cone $K$ is said to be proper if the following hold:

1. $Solid: int(K) \neq \emptyset$
2. $Closed: K$ is closed.
3. $Pointed: x\in K \land -x \in K \implies x = 0$

Definition Let $F \subset K \subset \mathbb{R}^n$ be pointed closed cones. $F$ is said to be a face of $K$ if $$x \in F,y\in K, x - y \in K \implies y \in F$$

Lemma If $F$ is a face of $K$, then $F = K \cap \{x - y \ | \ x,y \in F \}$. If $F \notin \{ \{0\}, K\}$, then $F \subset \partial K$.

Theorem Let $A \in \mathbb{R}^{n,n}$ and $K$ a proper cone. Suppose that $AK \subset K$, i.e., $K$ is invariant under $A$. Then $K$ contains an eigenvector of $A$.

Definition Let $A \in \mathbb{R}^{n,n}$ and $K$ a proper cone. $A$ is said to be $K$-irreducible if $K$ is invariant under $A$ and the only faces of $K$ that are invariant under $A$ are $\{0\}$ and $K$.

---

This is the theorem. I only show the "if" part of the proof, since I understand the other one.

Theorem A matrix $A$ is $K$-irreducible if and only if no eigenvector lies on $\partial K$.

Proof (If:) Suppose $F$ is a nontrivial face of $K$. $F$ is a proper cone in $F - F$. Let $A_F$ be the restriction of $A$ to $F - F$. By above theorem, $A_F$ has an eigenvector $x\in F$, but $x$ is also an eigenvector of $A$ and $x \in \partial K \square$.

Questions

1. What on earth does "$F$ is a proper cone in $F - F$" mean? I know that $F - F = \{ x - y | x,y\in F\}$. Is it because $F - F$ is a subspace?
2. Where is the contradiction in the proof? I mean, the last statement says that $x \in \partial K$, how is this a contradiction?