I'm trying to prove that if for every valuation $V : F \to \mathcal O (\mathbb Q)$, we have $V, \Gamma \models A$, (defined by $V(\Gamma) \subseteq V(A)$), then $\Gamma \vdash A$ (completeness of topological semantics with respect to $\mathbb Q$). I'm assured this is true for propositional logic with $\mathbb Q$ or $\mathbb R$ or any product of them (Lectures on the Curry-Howard Isomorphism, theorem 2.4.11).
I tried induction on complexity of formulas, but I hit a problem with disjunction:
If $V(\Gamma) \subseteq V(A) \cup V(B)$, then $\Gamma \vdash A \vee B $. I'd want to use the inductive hypothesis to get either $\Gamma \vdash A$ or $\Gamma \vdash B$ and derive the disjunction, but you can't get either $V(\Gamma) \subseteq V(A)$ or $V(\Gamma) \subseteq V(B)$. If $\Gamma = \emptyset $, then we want $\mathbb Q \subset V(A)$ or $\mathbb Q \subset V(B)$, but both are false if $V(A) = (0, \infty)$ and $V(B) = (-\infty, 1)$.
EDIT: I'm now quite certain that this sort of proof won't work. I'm pretty sure the way other proofs go is by contrapositive, constructing a valuation $V$ such that if $\Gamma \not \vdash A$, then $V(\Gamma) \not \subseteq V(A)$, possibly depending on $\Gamma$ and $A$. I haven't yet attempted this proof but it looks difficult and a reference would still be useful.
Perhaps this proof on induction on complexity doesn't work. I'd like to know how to do a direct proof of this, or to have a reference for topological semantics. I don't like Lectures on the Curry-Howard Isomorphism because it's left as a rather long exercise which goes through Kripke semantics, and I want to do it directly. A proof for $\mathbb R$ would be fine too.