Theorem 9.6.4 (Hahn-Banach, First Geometric Form [Theorem 1.6, Bre10]). Let $E$ be a TVS over $\real$, $A, B \subset E$ be non-empty convex sets, such that $A \cap B = \emptyset$ and $A$ is open, then there exists $\phi \in E^{*} \setminus \bracs{0}$ and $\alpha \in \real$ with $A \subset \bracs{\phi \le \alpha}$ and $B \subset \bracs{\phi \ge \alpha}$.
Proof. Let $C = A - B$, then $C \subset E$ is an open convex set by Lemma 9.1.3. Since $A \cap B = \emptyset$, $0 \not\in C$.
By Lemma 9.6.3, there exists $\phi \in E^{*}$ such that $C \subset \bracs{\phi < 0}$. In which case, for any $a \in A$ and $b \in B$, $\dpb{a - b, \phi}{E}< 0$ implies that $\dpb{a, \phi}{E}< \dpb{b, \phi}{E}$. Let $\alpha \in [\sup_{a \in A}\dpb{a, \phi}{E}, \inf_{b \in B}\dpb{b, \phi}{E}]$, then $A \subset \bracs{\phi \le \alpha}$ and $B \subset \bracs{\phi \ge \alpha}$.$\square$