Group orderable iff all its finitely-generated subgroups are orderable I want to proof this specifically using the Compactness Theorem from propositional logic (this is an exercise from Model Theory, Hodges).
$G$ orderable means there is a total ordering s. t. for all $g,h$,  $g\leq h$ implies $gk\leq hk$ and $kg\leq kh$ for all $k$.
My idea is to take the following set of formulae:
$$\{p_{gh}\lor p_{hg}|g,h\in G\}\cup \{(p_{gh}\land p_{hi})\implies p_{gi}|g,h,i\in G\}\cup \{p_{gh}\implies\neg p_{hg}|g\neq h\in G\}\cup \{p_{gh}\implies(p_{ab}\land p_{cd})|a=kg, b=kh, c=gk, d=hk; g,h,k\in G\}$$
Models of this set should precisely be total orderings that meet the orderable requirement if I'm not mistaken? But I struggle to use compactness from here. All finitely-generated subgroups being orderable doesn't correspond to all subsets of above formulae-set having a model, or does it?

To apply compactness, you don't need all subsets of your set to have models; you just need all finite subsets to have models.  A finite subset $S$ of your set involves only finitely many of the propositional variables $p_{gh}$ which in turn involve only finitely many elements of the group $G$.  The subgroup $H$ of $G$ generated by these elements is finitely generated, so by assumption $H$ is orderable.  Picking an order on $H$ and using it to determine the truth of $p_{gh}$ whenever $g,h\in H$ will then make all the formulas in $S$ true: that is, it will give a model of $S$.  Thus every finite subset of your set has a model, so by compactness the whole set has a model.
