Is it vacuously true?
Prove or disprove that for every theory $T$, if $T$ is not satisfiable then for every $\phi$, $T \vdash \phi$.
If $T$ is not satisfiable, then there is no structure $\mathcal M$ so that $\mathcal M \vDash T$, so it just seem vacuously true that for every $\phi$, $T \vdash \phi$.
On
This statement is not quite vacuous as it seems.
There is a difference between $\vdash$, which means "proves" and $\models$ which means "logically entails". The former is about the existence of a proof, the latter is about the truth value in models.
The completeness theorem for first-order logic, however, says that the two are very related. How related? They happen and fail simultaneously. But it does not meant that they are the same, or that this happens in logic which is not first-order logic.
If $T$ is not satisfiable, then the completeness theorem says that $T$ is inconsistent, therefore $T$ proves a statement and its negation, and from the explosion principle we know that a contradiction proves everything, so $T$ proves everything.
Or, alternatively, since $T$ is not satisfiable, it vacuously holds that $T\models\phi$, and by the completeness theorem $T\vdash\phi$. But there is a point where you use completeness, and it's a nontrivial point.
Your statement is more known in the non-contrapositive form of the model existence lemma:
Its proof from is easily found in most logic textbooks, such as van Dalen, Goldrei and others.
Note, however, that it's not a matter of the statement being vacuosly true. The point is because if $T$ has no model, some subset of is not satisfiable. This means that some subset of has the form $\{\varphi,\neg\varphi\}$, so that from a contradiction we trivialize $T$, that is, for every $\psi \in T$, $T \vdash \psi$. This is not supposed to be a proof though.