I'm trying to show that if a set of sentences $\Sigma$ is such that for every valuation $v:A \rightarrow 2$, there is some $p\in\Sigma$ with $v^*(p)=1$, then there is are $p_1,\ldots,p_n\in\Sigma$ such that $p_1\vee \cdots \vee p_n$ is a tautology.
Like the title suggests, I suspect that compactness is somehow involved in the proof, but I don't have a clear sense of how to get started.
Consider the set $ \overline \Sigma = \{\neg p \mid p \in \Sigma\}$. By hypothesis, $\overline \Sigma$ has no model: for every valuation $v$, there's $\neg p = q \in \overline \Sigma$ with $v(q) = 0$ (for some $p \in \Sigma$ for which $v(p) = 1$). By the compactness theorem in contrapositive form, some finite subset $S_0 = \{q_0, \dots, q_n\} = \{\neg p_0, \dots, \neg p_n\} \subseteq \overline \Sigma$ has no model. That is, $$(\neg p_0 \wedge \dots \wedge \neg p_n)$$ is false in every model (i.e. for every valuation). So its negation, $$p_0 \vee \dots \vee p_n$$ is true in every model, i.e. it's a tautology.