An "open-strict" Version of Hahn Banach Separation Theorem? Is the following statement true?
Let $X$ be a real linear space, $A,B \subset X$ two disjoint convex sets with the following "algebraic openness" property: Every $x \in A$ is an internal point of A, and so is every point of B and internal point of B. Then there exists a linear functional $f:X \rightarrow \mathbb{R}$ and $t \in \mathbb{R}$ such that for all $x \in A, y \in B$ we have 
$$ f(x) < t < f(y)$$
A similar version (but in the context of TVS and one of the sets is open) is given in Wikipedia, Note that it only admits a "half strict" separation. I believe that it is still true when topological openness is replaced by the "algebraic openness" definition given here, but the question is if I assume both sets are "algebraically open", is it true that I can get a strict separation from both sides? Any help is appreciated. 
 A: IMPORTANT EDIT: A theorem related to the question stated in the OP can be found here. Theorem 4 states: If $A$ and $B$ are disjoint convex sets in $X$ and $A$ has an internal point, then $A$ and $B$ can be (weakly) separated. That is, there exists $f:X\rightarrow\mathbb{R}$ such that $$\sup_{a\in A}f(a)\leq\inf_{b\in B}f(b).$$
EDIT 2: We can use this theorem to answer the question in the OP. This follows from the following proposition.
Proposition: Let $f:X\to\mathbb{R}$ be linear and non-zero. Then for any $A$ obeying the "algebraic openness" property, we have that $f(A)$ is open.
Proof: Let $t\in f(A)$. So $f(a)=t$ for some $a\in A$. Because $f$ is non-zero, we find some $x\in X$ such that $f(x)>0$. By the "algebraic openness" property of $A$, there exists $\varepsilon>0$ such that $a+(-\varepsilon,\varepsilon)\cdot x\subset A$. Hence, $(t-\varepsilon f(x),t+\varepsilon f(x))\subseteq f(A)$, so $f(A)$ is open.
Combining the two results, we have for all $a\in A$ and $b\in B$ that $$f(a)<\sup_{\alpha\in A}f(\alpha)\leq\inf_{\beta\in B}f(\beta)<f(b).$$
ORIGINAL ANSWER: The set of all sets with your "algebraic openness" property makes $X$ a topological vector space. Hence, if $A$ and $B$ are disjoint, convex and open in this topology, then there exists a continuous linear functional $\phi$ and a constant $s\in\mathbb{R}$ such that $\phi(a)<s\leq\phi(b)$ for all $a\in A$ and $b\in B$. But there also exists a continuous linear functional $\psi$ and a constant $t\in\mathbb{R}$ such that $\psi(b)<t\leq\psi(a)$ for all $b\in B$ and $a\in A$. Then $f:=\phi-\psi$ is a continuous linear functional such that $f(a)<s-t<f(b)$ holds for all $a\in A$ and $b\in B$.
EDIT 3: We now know that the "algebraic openness" property does not define a topological vector space. Do algebraically open sets define a vector space topology?
