I little confused on compactness of propositional logic.
So propositional logic has the property of being compact, that is to say, given a set of formulas $\mathcal F$, then $\mathcal F$ is satisfiable if and only if every finite subset of $\mathcal F$ is satisfiable.
But what about the following: If $\mathcal F = \{F_0, F_1, \ldots \}$ such that
$F_0$ = there are finitly many things in the universe.
$F_1$ = there is at least one thing in the universe.
$F_2$ = there are at least two things in the universe.
.
.
.
$F_n$ = there are at least $n$ things in the universe.
Clearly $\mathcal F$ is not satisfiable even though every finite subset is satisfiable. Does that mean $\mathcal F$ is not in propositional logic? Why?
Because you cannot express $F_0$ in propositional logic.
In fact, in propositional logic, there's no real universe of objects - models of propositional languages are simply assignments of truth values to atomic formulas - so you can't really speak about "how many things there are" at all.
You can do so, however, in predicate logic. For example, "There are at least $n$ things" can be expressed as $$ \varphi_n = \exists x_1 \ldots \exists x_n \,:\, x_1 \neq x_2 \land \ldots \land x_1 \neq x_n \land x_2 \neq x_3 \land \ldots \land x_2 \neq x_n \land \ldots $$ and "There are at most $n$ things as $$ \theta_n = \lnot \varphi_{n+1} $$
But even in predicate logic, how would you express "There are finitely many things"? You can't use $$ \theta_1 \lor \theta_2 \lor \ldots $$ because that formula would have infinite length.
In fact you can extend the reasoning in your question, by applying it to first-order predicate logic instead, to rigorosly prove that $F_0$ isn't expressible in first-order predicate logic.