Equivalent definitions for "$f$ separates $A$ and $B$" 
The book defines two other statements as equivalents for $f$ separating two sets $A$ and $B$ underlined in the above-text with orange and green. I cannot represent a proof for neither cases and neither direction in each case. Please help, thanks!
 A: Let (i) be the initial definition, (ii) be the definition that comes after the orange underline and (iii) be the definition that comes after the green underline. 
Then, as a hint, try showing that (i)$\Leftrightarrow$(ii)$\Rightarrow$(iii)$\Rightarrow$(i).
For (i)$\Rightarrow$(ii), use the linearity of $f$.
For (ii)$\Rightarrow$(i), use again linearity of $f$ by using $f(x-y)=f(x)-f(y)$ and $f(0)=0$.
For (ii)$\Rightarrow$(iii), what happens if $f$ separates $A$ and $B$ and $A'\subset A$ ?
For (iii)$\Rightarrow$(i), If $A=B$ and they are not a singleton, consider $x\neq y\in A$ as $x_0$ to conclude. If not, we can assume wlog that there exists $x_0\in A\backslash B$. If there exists only one element in $A\backslash B$, it goes back to a similar argument than for the first case.
Finally, if $|A\backslash B|\geq 2$, consider $x\neq y\in A\backslash B$ to be able to conclude.
A: First, let $A$ and $B$ be subsets of $X$ and $f$ be a linear functional on $X$.

We say that $f$ \emph{ separates} $A$ and $B$ if  there is $c \in \Bbb R$, such that:
either, for all $x \in A$ and $y \in B$, $f(x)\geq c$ and $f(y)\leq c$, or, for all $x \in A$ and $y \in B$, $f(x)\leq c$ and $f(y)\geq c$.

The "or" is needed to ensure that if $f$ separates $A$ and $B$, then $f$ separates $B$ and $A$.
Now, let us prove that

$f$ separates $A$ and $B$  if and only if  $f$ separates $\{0\}$ and $A-B$.

Proof:
$(\Rightarrow)$ If $f$ separates $A$ and $B$, then there is $c \in \Bbb R$, such that:
either, for all $x \in A$ and $y \in B$, $f(x)\geq c$ and $f(y)\leq c$, or, for all $x \in A$ and $y \in B$, $f(x)\leq c$ and $f(y)\geq c$.
So, for all $z \ A-B$, $f(z) \geq 0$ or $f(z) \leq 0$.
Since $f(0)=0$, we have that, for all $z \in \ A-B$ and $w\in \{0\}$, $f(z) \geq 0$ and $f(w) \leq 0$ or, for all $z \in \ A-B$ and $w\in \{0\}$, $f(z) \leq 0$ and $f(w)\geq 0$.
So, $f$ separates $\{0\}$ and $A-B$.
$(\Leftarrow)$ Suppose that $f$ separates $\{0\}$ and $A-B$. Then, for all $z \in \ A-B$, $f(z) \geq 0$ or $f(z) \leq 0$.
Suppose for all $z \in \ A-B$, $f(z) \geq 0$. Then, it means that, for all $x \in A$ and $y \in B$, $f(x-y) \geq 0$. It means that, for all $x \in A$ and $y \in B$,   $f(x) \geq f(y)$. So, $\inf f(A) \geq \sup f(B)$. So, there is $c \in \Bbb R$, such that $\sup f(B) \leq c \leq \inf f(A)$. It follows immediately that $f$ separates $A$ and $B$.
Suppose for all $z \in \ A-B$, $f(z) \leq 0$. The proof is completely similar to the previous paragraph.
$\square$
Now let us prove that

$f$ separates $A$ and $B$  if and only if, for every $x_0 \in X$, $f$ separates $A-\{x_0\}$ and $B-\{x_0\}$.

Note that $A-\{x_0\}=\{x-x_0 : x \in A\}$ and $B-\{x_0\}=\{y-x_0 : y \in B\}$.
Proof:
$f$ separates $A$ and $B$ means that there is $c \in \Bbb R$, such that:
either, for all $x \in A$ and $y \in B$, $f(x)\geq c$ and $f(y)\leq c$, or, for all $x \in A$ and $y \in B$, $f(x)\leq c$ and $f(y)\geq c$.
It is equivalent to: either, for all $x -x_0 \in A-\{x_0\}$ and $y -x_0  \in B-\{x_0\}$, $f(x-x_0 )\geq c-f(x_0)$ and $f(y-x_0 )\leq c-f(x_0)$, or, for all $x-x_0 \in A-\{x_0\}$ and $y-x_0 \in B-\{x_0\}$, $f(x-x_0 )\leq c-f(x_0)$ and $f(y-x_0 )\geq c-f(x_0)$. It means, $f$ separates $A-\{x_0\}$ and $B-\{x_0\}$.
$\square$
Remark: Note that $A$ and $B$ don't need to be convex subsets for the results above.
