 Solution

Suppose $S$ and $R$ are model companions of $T$. Their universal consequences of a theory agree, so the diagram embedding criterion for universal theories permits embeddings in both directions between their model classes.

Starting from any $M_0\models S$, alternately take such extensions, identifying each model with its image under the embedding:Since $S$ is model-complete, $M_i\preccurlyeq M_{i+1}$; since $R$ is model-complete, $N_i\preccurlyeq N_{i+1}$. The two subsequences have the same union $U$. The elementary chain theorem givesEvery axiom of $R$, as a sentence true in $U$, is therefore true in $M_0$. Thus every model of $S$ is a model of $R$. Reversing their roles proves the converse. The model companion is unique up to logical equivalence. If $T$ is inconsistent, its only possible companion is likewise inconsistent, so the same uniqueness conclusion holds.