On page 35 in Ziegler, Tent: Model Theory it is stated:
Let $T$ be an $L$-theory. It follows from 3.1.2. that the models of $T_{\forall}$ are the substructures of models of $T$.
with
Lemma 3.1.2. Let $T$ be a theory, $\mathfrak{A}$ a structure and $\Delta$ a set of formulas closed under existential quantification, conjunction and substitution of variables. Then the following are equivalent:
a) All sentences $\varphi\in\Delta$ which are true in $\mathfrak{A}$ are consistent with $T$.
b) There is a model $\mathfrak{B}\vDash T$ and a map $f:\mathfrak{A}\to_\Delta \mathfrak{B}$.
I don't see why it follows directly from 3.1.2.
I try to argue by this, but I am not sure if it makes much sense:
Assume that $T$ is consistent. As $T_\forall=\{\psi_\forall|\psi_\forall\text{ universal and } T\vdash\psi_\forall\}$ we have that $T_\forall\subseteq\text{Ded}(T)$ (deductive closure) and therefore $T_\forall\cup\text{Ded}(T)$ and also $T_{\forall}\cup T$ are consistent. Now, pick a model $\mathfrak{A}_\forall\vDash T_\forall$. By the equivalence of 3.1.2. we have that there is a model $\mathfrak{B}\vDash T$ with a map $f:\mathfrak{A}_{\forall}\to_{T_\forall}\mathfrak{B}$. I see no reason why I can pick $f$ to be the inclusion map.
Conversely, I don't see why if I have $\mathfrak{B}\vDash T$ and the inclusion $\iota:\mathfrak{A}\to_{T_\forall}\mathfrak{B}$, that then $\mathfrak{A}\vDash T_\forall$.
I think the authors' intention is to apply 3.1.2 with $\Delta$ being the set of existential formulas. Then the proof of the alleged consequence would go like this. Suppose $\mathfrak A\models T_\forall$. Then I claim clause (a) of 3.1.2 is true. Indeed, if $\phi$ is an existential sentence true in $\mathfrak A$, then $\neg\phi$ is (up to logical equivalence) a universal sentence that is false in $\mathfrak A$ and therefore is not in $T_\forall$. So $\phi$ is consistent with $T$. By 3.1.2, we infer that clause (b) is also true, so we have $\mathfrak B\models T$ and $f:\mathfrak A\to_\Delta\mathfrak B$.
Now we need another (standard but easy to get wrong) maneuver to get $f$ to be an inclusion. As it stands, $f$ is an isomorphism from $\mathfrak A$ to a substructure $\mathfrak A'$ of $\mathfrak B$ (because all atomic formulas and their negations are in $\Delta$). The idea is to modify $\mathfrak B$ by replacing this copy $\mathfrak A'$ of $\mathfrak A$ with $\mathfrak A$ itself. That way you get an isomorphic copy of $\mathfrak B$ (hence still a model of $T$) that contains literally $\mathfrak A$ rather than a copy $\mathfrak A'$ of it. The problem (the reason it's easy to get wrong) is that, if we're unlucky, $\mathfrak B$ might already contain some elements of $\mathfrak A$, and not in the place where we want them; then if we just replace $\mathfrak A'$ with $\mathfrak A$ we'll end up with some elements of $\mathfrak A$ trying to be in two different places in $\mathfrak B$. Fortunately, the cure for this problem is easy: First replace $\mathfrak B$ by an isomorphic copy that is disjoint from $\mathfrak A$, and then proceed as above.
Finally, to show the converse, that all substructures of models of $T$ are models of $T_\forall$, prove, by induction on universal formulas $\theta$, that, if $\mathfrak A$ is a substructure of $\mathfrak B$ and if some assignment of values for variables in $\mathfrak A$ satisfies $\theta$ in $\mathfrak B$, then it also satisfies $\theta$ in $\mathfrak A$. The only nontrivial step in the induction is the universal quantifier step, and there you just use that all elements of $\mathfrak A$ (that might serve as counterexamples) are also in $\mathfrak B$ (and therefore cannot be counterexamples).