Theorem 5.30 (Separation Theorem)
Suppose that $X$ is a real or complex normed space and $A,B\subset X$ are non-empty, disjoint, convex sets.
(a) If $A$ is open then there exists $f\in X^\prime$ and $\gamma\in\mathbb{R}$ such that$$\Re\text{e}f(a)<\gamma\leq\Re\text{e}f(b),\quad a\in A,\:b\in B.$$(5.11)
Proof
By Exercise 5.10, it suffices to consider the real case (of course, if $X$ is real weomit $\Re$e in (5.11)and (5.12)).
(a) Choose $a_0\in A,b_0\in B$, and let $w_0=b_0-a_0$, and $C=w_0+A-B.$ Then $C$ is a convex, open set containing 0, so the Minkowski functional $p_C$ is well-defned and sublinear. Also, since $A$ and $B$ are disjoint, $w_0\not\in C$, so by (5.9), $p_{C}( w_{0}) \geq 1.$
Let$W=$Sp$\left\{w_{0}\right\}$,and define a linear functional $f_{W}$ on $W$ by $f_{W}(\alpha w_{0})=\alpha$,$\alpha\in\mathbb{R}$. If $\alpha\geq0$ then$$f_W(\alpha w_0)\leq\alpha p_C(w_0)=p_C(\alpha w_0),$$while if $\alpha<0$ then$$f_W(\alpha w_0)<0\leq p_C(\alpha w_0).$$Thus $f_{W}$ satisfies (5.3), and so, by Theorem 5.13, $f_{W}$ has an extension $f$ on $X$
which satisfes (5.4). Combining this with (5.5) and (5.10) shows that $f\in X^\prime.$
Now, for any $a\in A,b\in B$, we have $w_{0}+a-b\in C$, so that$$\begin{aligned}1+f(a)-f(b)=f(w_0+a-b)\leq p(w_0+a-b)<1,\end{aligned}$$by Lemma 5.29. Hence, defining $\gamma=\inf\{f(b):b\in B\}$, this yields$$f(a)\leq\gamma\leq f(b),\quad a\in A,\:b\in B.$$(5.13)
To obtain (5.11) from this, suppose that there is $a\in A$ such that $f(a)=\gamma.$ Since $A$ is open there exists sufficiently small $\delta>0$ such that $a+\delta w_0\in A$, and hence,$$f(a+\delta w_0)=f(a)+\delta f_W(w_0)=\gamma+\delta>\gamma,$$which contradicts (5.13), and so completes the proof of (5.11)
Question: Since $A$ is open there exists sufficiently small $\delta>0$ such that $a+\delta w_0\in A$. I don't understand why $a+\delta w_0\in A$? If $w_0\in A$, then I can understand. But it is not.