The following problem is from the book "A Beginner's Guide to Mathematical Logic" by Raymond M. Smullyan in the context of preparing a (second) proof of the compactness theorem for propositional logic (slightly reworded to use unsigned instead of signed formulas):
For any set $S$ of formulas and any formula $X$, if some finite subset of $S \cup \{X\}$ is unsatisfiable, and if some finite subset of $S \cup \{\neg X\}$ is unsatisfiable, then some finite subset of $S$ is unsatisfiable.
My proof goes like this:
Proof. Let $I$ be an arbitrary interpretation and $S_1 \subseteq S \cup \{X\}$ and $S_2 \subseteq S \cup \{\neg X\}$ some unsatisfiable finite subsets. Then either $I$ satisfies $X$ or $I$ satisfies $\neg X$. If $I$ satisfies $X$, then $I$ cannot satisfy $S_1 \setminus \{X\}$, for $S_1$ is unsatisfiable. If $I$ satisfies $\neg X$, then $I$ cannot satisfy $S_2 \setminus \{\neg X\}$, for $S_2$ is unsatisfiable. Thus $(S_1 \setminus \{X\}) \cup (S_2 \setminus \{\neg X\})$ is an unsatisfiable finite subset of $S$. $\blacksquare$
Now I have tried an alternative proof of the contrapositive, i.e.:
For any set $S$ of formulas and any formula $X$, if all finite subsets of $S$ are satisfiable, then either all finite subsets of $S \cup \{X\}$ are satisfiable, or all finite subsets of $S \cup \{\neg X\}$ are satisfiable.
I began as follows:
Proof. Let $S_1 \subseteq S \cup \{X\}$ and $S_2 \subseteq S \cup \{\neg X\}$ be finite subsets. If $X$ is unsatisfiable, then $\neg X$ is valid and since all finite subsets of $S$ are satisfiable, $S_2$ is satisfiable as well, hence all finite subsets of $S \cup \{\neg X\}$ are satisfiable. If $X$ is satisfiable, ...
At this point I don't know how to continue, i.e. how to show that if both $X$ and $S_1 \setminus \{X\}$ are satisfiable, then they necessarily have at least one interpretation in common that satisfies both, such that $S_1$ is satisfiable and hence I could conclude that all finite subsets of $S \cup \{X\}$ are satisfiable. Any hints on how the proof could proceed, if it can?