 Solution

If $T$ is model-complete, the inclusion of a substructure of a first-order structure $\mathcal{A}\subseteq \mathcal{B}$, both first-order models of $T$, is a structure embedding and hence elementary. Therefore $\mathcal{A}⪯\mathcal{B}$.

Conversely, assume every such inclusion is elementary, and take any structure embedding $f:\mathcal{A}\to \mathcal{B}$ between first-order models of $T$. Its image $f[A]$, with the induced first-order structure, is a substructure of $\mathcal{B}$ isomorphic to $\mathcal{A}$, so it is itself a first-order model of $T$. By hypothesis $f[A]⪯\mathcal{B}$. The isomorphism $\mathcal{A}\to f[A]$ preserves all first-order formulas, and composing it with this elementary inclusion shows that $f$ is an elementary embedding. Thus $\boxed{T\text{ is model-complete}⟺(\mathcal{A}\subseteq \mathcal{B}\models T\Rightarrow \mathcal{A}⪯\mathcal{B})}$.