Proof.

1. (1)

   Let $E=A-B$ then $0\nin E$. $E$ is convex. $\lam\in [0,1]$ and $z_{1},z_{2}\in E$

   \begin{align*}(1-\lam)z_{1}+\lam z_{2}=(1-\lam)(x_{1}-y_{1})+\lam (x_{2}-y_{2})=\braks{(1-\lam)(x_1)+\lam (x_2)}-\braks{(1-\lam)(y_1)+\lam (y_2)}\in E\end{align*}

   
   

   $E$ is closed. Let $\curl{z_k}_{k\in\N}\suf E$ such that $z_{k}=x_{k}-y_{k}\to \to z$. Since $A$ is compact there exists a subsequence $x_{k_m}\to x\in C$ then $y_{k_m}=x_{k_m}-z_{k_m}\to x-z=y\in B$ since $B$ is closed. Thus $z=x-(x-y)\in A-B=E$. By Theorem 13.1.25 $\exists a\in\R^{n}\del\curl{\vec{0}}$ and $\al'\in\R$ such that $\inner{z}{a}>\al'$ for all $z\in\E$. So

   \begin{align*}\inner{x}{a}>\inner{y}{a}+\al'&&\fall x\in A,y\in B\\ \UBC{\INF{x\in A}\inner{x}{a}}{I}\geq\UBC{\SUP{y\in B}\inner{y}{a}}{S}+\al'\end{align*}

   
   

   Let $\al\define \frac{1}{2}(I+S),\e\define \al'/2$ hence

   \begin{align*}\al-\e&=\frac{I+S-\al'}{2}\geq \frac{S+\al'+S-\al'}{2}=S\\ \al+\e&=\frac{I+S+\al'}{2}\leq \frac{I+I-\al'+\al'}{2}=I\end{align*}

   
   

   showing $A,B$ are strongly separated.

2. (2)

   ^[1]

$\square$