The general argument goes like so: suppose $\Sigma$ is finitely satisfiable, but not satisfiable. Therefore, $\Sigma$ is inconsistent, so $\Sigma \vdash \bot$. Then, by completeness, there exists a derivation from $\Sigma$ of $\bot$ and, since every proof is finite, there exists a derivation of $\bot$ from a finite subset. Done.
However, I've been thinking about the fact that I've never actually seen the completeness theorem proven for an infinite set $\Sigma$. If we try to prove it in sequent calculus, for example, we have to put $\Sigma$ into a sequent; we'd encode that as $\Sigma \implies \bot$ --- but that's not a sequent, as the antecedent is infinite. How is this resolved?
There is indeed some care needed here.
Consider second-order logic with full semantics, $\mathsf{SOL}$. Let $\vdash_0$ be the "finitary entailment relation" for $\mathsf{SOL}$: given a (possibly infinite) set of $\mathsf{SOL}$-sentences $\Gamma$ and a single $\mathsf{SOL}$-sentence $\varphi$ (for simplicity I'm only allowing single-sentence consequents), we set $\Gamma\vdash_0\varphi$ iff some finite $\Gamma_0\subseteq\Gamma$ has $\Gamma_0\models_\mathsf{SOL}\varphi$.
Trivially $\vdash_0$ is complete for finite sequents: if $\Gamma$ is finite then $\Gamma\vdash_0\varphi$ iff $\Gamma\models_\mathsf{SOL}\varphi$. However, we also know that $\models_\mathsf{SOL}$ is not compact, whereas $\vdash_0$ by definition is compact, so $\vdash_0$ is not complete on infinite sequents.
So what? Well, this shows that there is in general a gap between completeness for finitary sequents and full completeness in general logics. In the particular case of first-order logic each result is true, but there are settings where one result holds and the other doesn't. So in fact your original concern was justified: if we state the completeness theorem in its weak form, we do not trivially get the stronger form, let alone compactness.
Fortunately, this is easily addressed by looking at the proof. Even if we don't want to modify our definition of sequents to accommodate infinite antecedents, we're still fine: supposing $\Gamma$ is a possibly-infinite set of $\mathsf{FOL}$-sentences no finite subset $\Gamma_0$ of which has $\Gamma_0\vdash\varphi$ being a valid sequent, we build via the usual Henkinization argument a model of $\Gamma\cup\{\neg\varphi\}$. So the only way we get in trouble is if we phrase the conclusion of this argument too blithely and treat it as a blackbox.