= Solution
The <soundness theorem for propositional logic> says that if $\Gamma\vdash t$, then every valuation satisfying every member of $\Gamma$ also satisfies $t$.
The proposed <function> need not respect the connectives. For example, if a primitive proposition $p$ is independent of $S$, then neither $p$ nor $\neg p$ is provable from $S$, so the definition gives
$$
v(p)=v(\neg p)=0.
$$
Order the consistent supersets of $S$ by inclusion. The union of a chain is consistent, since a finite proof of a contradiction would already use assumptions from one member of the chain. <Zorn lemma> therefore gives a maximal consistent extension $T$. It is deductively closed: if $T\vdash t$, adjoining $t$ preserves consistency, so maximality forces $t\in T$. Moreover, for every $t$, exactly one of $t$ and $\neg t$ belongs to $T$. They cannot both belong by consistency; if neither belonged, the inconsistency of both proper extensions would give $T\vdash\neg t$ and $T\vdash\neg\neg t$, again a contradiction. The usual induction on formulae now shows that
$$
v(t)=1\quad\Longleftrightarrow\quad t\in T
$$
defines a valuation satisfying $T$, and hence $S$.
Now suppose every finite subset of $S$ has a model. By soundness every finite subset is consistent. Any proof of a contradiction from $S$ uses only finitely many assumptions, so $S$ itself is consistent. Applying the preceding maximal-consistent-extension construction gives a model of $S$. This proves the <propositional compactness theorem>.
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