= Solution
If $B\models T$ and $B\subseteq A^*\equiv A$, every <universal sentence> of $A$ holds in $A^*$ and is preserved by passage to the <substructure of a first-order structure>[substructure] $B$. Thus $B$ witnesses satisfiability of the joint <first-order theory>[theory].
For <universal theory and substructures of elementarily equivalent models>, conversely let $B$ satisfy $T$ together with the universal <first-order theory>[theory] of $A$. Every <existential sentence> true in $B$ must be true in $A$: if an existential $\sigma$ were false in $A$, its universal <negation> would belong to $\operatorname{Th}_\forall(A)$ and hold in $B$, a <contradiction>. Hence $\operatorname{Th}_\exists(B)\subseteq\operatorname{Th}(A)$. In particular $\operatorname{Th}(A)\cup\operatorname{Th}_\exists(B)$ is satisfiable, with <first-order model>[model] $A$. Apply the preceding <structure embedding>[embedding] result to the <first-order theory>[theory] $\operatorname{Th}(A)$ and structure $B$. It supplies a structure $A^*\models\operatorname{Th}(A)$ containing $B$. Therefore \b[$A^*\equiv A$ and $B\subseteq A^*$], as required. <Elementary equivalence> here does not assert that $A$ itself contains a copy of $B$.
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