If $\Delta, \Delta' $ are satisfable sets of sentences while $\Delta \cup \Delta' $ is not satisfable then exists the such sentence $\phi $ that $\Delta \models \phi $ and $\Delta' \models \neg \phi$
Let's assume that there is no such $\phi$. Let take a finite subset: $\Delta_0 = \Delta \cup \Delta'$. Let's show that $\Delta_0$ is satisfable. Let's assume that is it unsatisfable. Then, no model of $\Delta'$ satisfies $\Delta_0$. It means that there exists a such $\phi$ that $\Delta \models \phi $ and $\Delta' \models \neg \phi$. What is a $\phi$? $\phi$ is just a conjuction of sentences from $\Delta$ ( $\Delta$ is finite). So $\Delta_0 $ was satisfable. By compactness theorem $\Delta \cup \Delta'$ is satisfable. Contraddiction.
How does that follow?
$\Delta \models \phi$. It is obvious because of construction of $\phi$.
$\Delta' \models \neg \phi$
No model of $\Delta'$ satisfies $\Delta_0$ where $\Delta_0 = (\Delta \cap \Delta_0) \cup (\Delta_0 \cap \Delta')$. It means that no model of $\Delta' \models \Delta_0 \cap \Delta \iff \text{ no model of } \Delta' \models \phi$. Therefore every model $M$ of $\Delta'$ $$M \models \neg \phi$$ and so $\Delta'\models \neg \phi$
But why does it follow from the fact that that $\Delta_0$ is satisfiable? Again, see my above counterexample for this:
Please note that at the beginnig I assume that there is no such $\phi$ as it is expected in the problem. Then, I want to prove that every finite subset is satisfable. So, I take $\Delta_0$ and assume that it is unsatisfable. Then I show that it follows that a such $\phi$ exists. So, contradiction. $\Delta_0$ must be satisfable.
... but \Delta_0 is still not satisfiable.
Yes, but note that there exists $\phi = P$
Right?