Let $\Gamma_1,\Gamma_2 \subseteq WFF.\;$ Prove: $\Gamma_1 \cup \Gamma_2$ is not satisfiable if and only if there exists $\alpha \in WFF$ such that $\Gamma_1 \vdash_{HPC} \alpha$ and $\Gamma_2 \vdash_{HPC} \lnot \alpha$
I succeeded to prove that if there is an $\alpha \in WFF$ such that $\Gamma_1 \vdash \alpha$ and $\Gamma_2 \vdash \lnot \alpha$, then $\Gamma_1 \cup \Gamma_2$ is not satisfiable. I'm having some trouble with the other direction. If either $\Gamma_1$ or $\Gamma_2$ is not satisfiable, then it's pretty easy. But what if both $\Gamma_1$ and $\Gamma_2$ are satisfiable?
If $\Gamma_1 \cup \Gamma_2$ is not satisfiable, but $\Gamma_1$ and $\Gamma_2$ both are, then you could consider the sentence that is the conjunction of all the sentences in $\Gamma_1$ (if $\Gamma_1$ is empty, then just consider any tautology). Let's call this statement $S_1$. Clearly $\Gamma_1 \vdash_{HPC} S_1$. Moreover, if it would be the case that $\Gamma_2 \not \vdash_{HPC} \neg S_1$, then that means that there is a valuation that sets all of the statements in $\Gamma_2$ to True, but $\neg S_1$ to False. But that means that that valuatin sets $S_1$ to True, and hence all sentences in $\Gamma_1$ to True, meaning that this valuatin would set all sentences in $\Gamma_1 \cup \Gamma_2$ to True, meaning that $\Gamma_1 \cup \Gamma_2$ would be satisfiable, whch contradicts the assumption. So, it must be true that $\Gamma_2 \vdash \neg S_1$. And so we have found our $\alpha$, as desired.