Let $V_1,V_2$ be convex subsets of a normed space $X$ with $V_1^\circ\neq\emptyset$ and $V_1^\circ\cap V_2 =\emptyset$. Then there exists $x'\in X'\setminus\{0\}$ such that $$\operatorname{Re} x'(v_1)\leq \operatorname{Re}x'(v_2)\quad\forall v_1\in V_1, v_2\in V_2$$
I started like this:
Since $V_1$ is convex, $V_1^\circ$ is convex, too. Hence by Hahn-Banach's separation theorem there exists $x'\in X'$ such that $\operatorname{Re} x'(v_1)\leq \operatorname{Re}x'(v_2)$ for all $v_1\in V_1^\circ$ and $v_2\in V_2$.
My idea was now to use Hahn-Banachs extension theorem: Since $V_1^\circ\subset V_1$ there exists an extension $\tilde x'$ such that $\tilde x'|_{V_1^\circ}=x'$. But my problem is I don't know how to conclude from this that $\operatorname{Re}\tilde x'(v_1)\leq \operatorname{Re} \tilde x'(v_2)$ for all $v_1\in V_1$, $v_2\in V_2$ and why $\tilde x'\neq 0$.
How can I argue here?