Prove that there exists separator between two sets of sentences FO. Let $\Delta$ and $\Gamma$ be such sets of formulas that for every $M$ either $M \models \Delta $ or $M \models \Gamma$. Prove that there exists a such $\phi$ that $\Delta \models \phi \text{ and } \Gamma \cup \{\phi\} $ is inconsistent.
My solution:
Let assume that there is no such $\phi$. 
Let's take a such $\psi $ that $\Delta \models \psi$.  ( We can consider that there exists such sentence because if it doesn't that means what $\Delta$ must be empty and it simpleto show contradiction.
So, for every  $\psi \in \Delta $ ( $\Delta \models \psi$) $\Gamma \cup \psi$ is consistent. So, $\Delta \cup \Gamma$ is consistent. So there is a such model $M$ that $M \models \Delta, M \models \Gamma$ . Contradiction.
Where am I wrong? I've seen solution using compactness theorem but I cannot recall myself it. So, I suppose that I am wrong but I cannot see why. Please explain :).
 A: As pointed in the comments, the key to this problem is realizing that $\Delta\cup\Gamma$ is consistent if we suppose there is no such $\phi$.
This is because of compactness. If either $\Delta$ or $\Gamma$ is inconsistent, then trivially there is such a $\phi$. Otherwise, if $S$ is a finite subset of $\Delta\cup\Gamma$ then either $S\subset \Delta$ or $S\subset \Gamma$, in which case $S$ is consistent by virtue of $\Delta$ and $\Gamma$ being consistent themselves.
At last, it might be that $S$ has a mixture of sentences of $\Delta$ and of $\Gamma$. Let $S_\Delta = S\cap\Delta$ and $S_\Gamma = S\cap \Gamma$. If such an $S$ is inconsistent, then we have that $\wedge S_\Delta \implies \neg \wedge S_\Gamma$. As $S_\Delta \subset \Delta$, then $\Delta \implies \neg \wedge S_\Gamma$, while $\Gamma \implies \wedge S_\Gamma$. That contradicts the hypothesis, so $S$ must be consistent.
Therefore, we have just shown that every finite subset of $\Delta\cup\Gamma$ is consistent, and therefore due to compactness the whole set of sentences is consistent. $\square$
