If a set of sentences $\Delta$ is satisfiable and $\Delta \subseteq \Gamma$, then $\Gamma$ must also be satisfiable. Show that this does not hold. Let $\Delta$ and $\Gamma$ be sets of sentences. Definition of satisfiable: If a set of sentences is satisfiable, then it has model.
If a set of sentences $\Delta$ is satisfiable and $\Delta \subseteq \Gamma$, then $\Gamma$ must also be satisfiable. Show that this does not hold.
Proof: There could be a model $\mathcal{M}$ s.t. $\mathcal{M} \models \Delta$ but $\mathcal{M} \not\models \Gamma$.
How's this for a proof?
For instance $\mathcal{M}$ could be s.t. for every sentence $\phi \in \Delta$, $\mathcal{M} \models \phi$. But because $\Delta \subseteq \Gamma$ it means there are sentences $\psi \in \Gamma$ and $\psi \not\in \Delta$. So now for every sentence $\psi \not\in \Delta$, but $\psi \in \Gamma$, $\mathcal{M} \not\models \psi$. Therefore, it's not the case that $\mathcal{M} \models \Gamma$, so $\mathcal{M}$ is a counterexample.

Assume $\Delta$ is satisfiable.  Choose any sentence $\psi \in \Delta$, and let $\Gamma = \Delta \cup \{ \lnot \psi \}$.
