Let $X$ be a real topological vector space, and let $A$ and $B$ be nonempty, disjoint, convex subsets of $X$.
S1. If $A$ is open, then there exists $c \in \mathbb R$ and a nonzero, linear, continuous, real-valued function $f$ on $X$ such that $f(a) \leq c$ for all $a \in A$ and $f(b) \geq c$ for all $b \in B$.
S2. If $A$ has nonempty interior, then there exists $c \in \mathbb R$ and a nonzero, linear, continuous, real-valued function $f$ on $X$ such that $f(a) \leq c$ for all $a \in A$ and $f(b) \geq c$ for all $b \in B$.
Both S1 and S2 are true, but I am wondering...
Question. Is there an easy way to see that S1 implies S2?
If S1 holds, and $A$ has nonempty interior, then we can find a nonzero, linear, continuous function $f$ that separates $B$ from int$A$. But I don't see that $f$ necessarily separates $B$ from all of $A$, nor do I see how to modify $f$ to make this the case. Any hints are appreciated.
Added. Does the following argument work?
As above, if S1 holds, then there is a nonzero, linear, continuous $f$ that separates $B$ and int$A$. Since $f$ is continuous, it also separates $B$ from cl(int$A$), the closure of int$A$. But in a topological vector space, cl(int$C$) = cl$C$ for any convex $C$, and so $f$ separates $B$ from cl$A$. It follows that $f$ separates $B$ and $A$.