There's a proof of the theorem in Enderton's book wherein the second half serves as an exercise, stated as follows:
Let $\Delta$ be a set of wffs such that (i) every finite subset of $\Delta$ is satisfiable, and (ii) for every wff $\alpha,$ either $\alpha\in\Delta$ or $(\neg\alpha)\in\Delta.$ Define the truth assignment $v:$ $$v(A)= \begin{cases} T&\text{iff }A\in\Delta,\\ F&\text{iff }A\not\in\Delta \end{cases} $$ for each sentence symbol (propositional constant) $A.$ Show that for every wff $\varphi,$ $\bar v(\varphi)=T$ iff $\varphi\in\Delta.$
Here's my take (which doesn't even require $\Delta$ to satisfy property (i)):
We'll proceed by induction on the degree of $\varphi\in\Delta$ (number of steps required to build $\varphi$ from the formation rules, in a language that contains only the connectives $\neg$ and $\rightarrow$). If $\varphi$ is an atomic formula, then by definition of $v,$ the result is obvious; $\bar v(\varphi)=T$ iff $\varphi\in\Delta.$
For the induction step, assume that $\bar v(\varphi)=T$ iff $\varphi\in\Delta$ for every formula $\varphi$ of arbitrary degree $n$ and we wish to prove the same for all $\varphi$ of degree $n+1.$ By the formation rules, there are only two possibilities:
Case: $\varphi=\neg\phi.$ Since $\Delta$ is complete, if $\neg\phi\in\Delta$ then $\phi\not\in\Delta,$ due to which by the induction hypothesis, $\bar v(\phi)=F$ therefore $\bar v(\neg\phi)=T.$ For the other direction, assume that $\bar v(\neg\phi)=T$ and suppose by way of contradiction that $\neg\phi\not\in\Delta.$ Again by completeness of $\Delta$ we have that $\phi\in\Delta$ which by the induction hypothesis means that $\bar v(\phi)=T.$ This is a contradiction.
Case: $\varphi=(\phi\rightarrow\psi).$ We proceed by contradiction -- given that $\varphi\in\Delta$ assume that $\bar v(\varphi)=F;$ in particular, that $\bar v(\phi)=T$ and $\bar v(\psi)=F.$ Since both $\phi$ and $\psi$ are a degree less than $\varphi,$ by the inductive hypothesis, we have that $\phi\in\Delta$ and $\neg\psi\in\Delta$ with which one can provide a proof of $\Delta\vdash\neg(\phi\rightarrow\psi)$ given a proof of the tautology $\vdash(\phi\rightarrow(\neg\psi\rightarrow\neg(\phi\rightarrow\psi))).$ This is a contradiction to the fact that $\Delta$ is complete since it can prove both $\varphi$ and $\neg\varphi.$
Lastly, we prove the other direction: assume that $\bar v(\varphi)=T.$ Now we have two subcases: $\bar v(\phi)=F$ or $\bar v(\psi)=T.$ By the induction hypothesis we have that $\neg\phi\in\Delta$ or $\psi\in\Delta$ with which, given a proof of $(\neg\phi\rightarrow(\phi\rightarrow\psi))$ and $(\psi\rightarrow(\phi\rightarrow\psi)),$ we always end up with $\Delta\vdash(\phi\rightarrow\psi).$ By completeness of $\Delta,$ we have that $(\phi\rightarrow\psi)\in\Delta$ and thus the proof is complete.
As you can see, no requirement for $\Delta$ to be finitely satisfiable at all. I don't see any mistake either, but if you do then I happily appreciate any correction. However, I'm still curious to know how the proof can be done requiring $\Delta$ to be finitely satisfiable. Any hints?