= Solution
Here $\forall_2$ denotes <universal-existential sentences>, with quantifier-free matrix and possibly empty blocks. Add constants for $A$ and consider
$$
\Sigma=T\cup\operatorname{Diag}(A)\cup
\{\neg\exists\mathbf y\,\theta(\mathbf c_a,\mathbf y):
\theta\text{ quantifier-free and }A\models\neg\exists\mathbf y\,\theta(\mathbf a,\mathbf y)\}.
$$
We show that $\Sigma$ is consistent. Suppose a finite subset is inconsistent. Combine its diagram facts as $\delta(\mathbf c)$ and its finitely many negative existential facts as $\neg\exists\mathbf y_j\,\theta_j(\mathbf c,\mathbf y_j)$. Replacing the new constants by a variable tuple gives the consequence
$$
T\models\forall\mathbf x\left(\delta(\mathbf x)\longrightarrow
\bigvee_{j=1}^m\exists\mathbf y_j\,\theta_j(\mathbf x,\mathbf y_j)\right).
$$
This is equivalent to a universal-<existential sentence>: move all the existential blocks to the front of the disjunction, using the usual nonempty-domain convention. If there are no negative existential facts, the consequence is simply $\forall\mathbf x\,\neg\delta$. By hypothesis $A$ satisfies the displayed consequence. Its named tuple satisfies $\delta$ but none of the existential alternatives, a contradiction.
A model $B$ of $\Sigma$ contains a diagram copy of $A$. Every <existential formula> false in $A$ is still false in $B$; every one true in $A$ is preserved by the embedding. Thus
$$
\boxed{A\preccurlyeq_1 B\models T.}
$$
Both preservation and reflection are required here; the atomic diagram alone would give only an ordinary embedding.
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