= 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:
$$
M_0\subseteq N_0\subseteq M_1\subseteq N_1\subseteq M_2\subseteq\cdots,
\qquad M_i\models S,\quad N_i\models R.
$$
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> gives
$$
U\models S\cup R,\qquad M_0\preccurlyeq U.
$$
Every 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. \b[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.
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