$\newcommand{\Mod}{\operatorname{Mod}}\newcommand{\Th}{\operatorname{Th}}$
Give counterexamples to $\Th(K_1 \cap K_2) \subseteq \Th(K_1)\cup \Th(K_2)$ and $\Mod(\Gamma \cap \Delta) \subseteq \Mod(\Gamma) \cup \Mod(\Delta)$
In this context, $K_1$ and $K_2$ are classes of $\Sigma$-structures and $\Gamma$ and $\Delta$ are sets of formulas and $$\Th(K) = \{ \varphi \in \operatorname{Sent}_\Sigma \mid \mathcal{A} \models \varphi, \, \forall \mathcal{A} \in K\}$$ and $$\Mod(\Phi) = \{ \mathcal{N} \mid \mathcal{N} \models \phi, \, \forall \phi \in \Phi\}$$ I have already shown that the inclusion "$\supseteq$" always holds and I have been trying to come up with a counterexample showing that the other inclusion, "$\subseteq$" does not hold in general, but failed at the attempt. Any hint on how to proceed would be highly appreciated. Thank you!
UPDATE In a new attempt to come up with something, let $\varphi = \exists x P(x)$, $\psi_1 = \exists x P(x) \rightarrow \exists x R(x)$ and $\psi_2 = \exists xP(x) \rightarrow \exists xQ(x)$. If our universe only has one element $a$, that is $A = \{ a \}$. Let $\Gamma = \{\varphi, \psi_1 \}$ and $\Delta = \{ \varphi, \psi_2 \}$. Then $\Gamma \cap \Delta = \{ \varphi \}$. Then, a model $\mathcal{N}$ for $\Gamma \cap \Delta$ can be constructed by letting $P^{\mathcal{N}}(a) = 1$ and $Q^{\mathcal{N}}(a) = 0$. However, $\mathcal{N} \not \models \psi_1$ and $\mathcal{N} \not \models \psi_2$, therefore, $\mathcal{N} \not \in \Mod(\Gamma) \cup \Mod(\Delta)$. Is this correct?