Given a standard definition of an abstract logic $\mathcal{L}_A$ (cfr. Barwise & Feferman, Model-theoretic Logics, Springer-Verlag, 1985), let $E_A(\sigma)$ be the class of $\mathcal{L}_A$-sentences for a signature $\sigma$. Two $\sigma$-structures are $\mathcal{L}_A$-equivalent iff they satisfy the same sentences in $E_A(\sigma)$.
The finite Robinson property is defined as follows: given two signatures $\sigma$ and $\sigma'$, for all $\varphi \in E_A(\sigma)$, $\varphi' \in E_A(\sigma')$ and $\Psi \subseteq E_A(\sigma \cap \sigma')$, if $\Psi \cup \{\varphi\}$ and $\Psi \cup \{\varphi'\}$ are both satisfiable and every two $(\sigma \cap \sigma')$-models of $\Psi$ are $\mathcal{L}_A$-equivalent, then $\Psi \cup \{\varphi, \varphi'\}$ is satisfiable.
Let $\mathcal{L}_{\omega_1\omega}$ the abstract logic obtained from first-order logic allowing countably-long disjunctions (hence also conjunctions). An exercise in Keisler's Model Theory for Infinitary Logic: Logic with Countable Conjunctions and Finite Quantifiers (North Holland Publishing Company, 1971) says that $\mathcal{L}_{\omega_1\omega}$ has finite Robinson property if $\Psi$ in the above definition is countable, otherwise not. So it exists an uncountable $\Psi$ which is a counterexample to the property. Any suggestion?