An usual exercise in introductions to model theory is to prove the following statement:
Let $T$ be a theory, and let $T_\forall$ be the set of all universal sentences that are consequences of $T$. Then $\mathbf A$ is a model of $T_\forall$ if, and only if, there exists a model $\mathbf B$ of $T$ with $\mathbf A\subseteq \mathbf B$.
The proof of this fact only uses that the theory $T\cup \Delta(\mathbf A)$ is consistent, where $\Delta(\mathbf A)$ is the atomic diagram of $\mathbf A$. Now I wonder if the analogue for existential consequences is true:
Let $T$ be a theory, and let $T_\exists$ be the set of all existential sentences that are consequences of $T$. Then $\mathbf A$ is a model of $T_\exists$ if, and only if, there exists a model $\mathbf B$ of $T$ with $\mathbf B\subseteq \mathbf A$.
I have no evidence that this holds, apart from the fact that the $\forall$- and $\exists$-preservation theorems relate in the same way than the previous statements do, and that the $\forall$-preservation theorem can be proved using the first statement above.
It is not true.
Consider the language $L$ with a single binary predicate and a model $N$ which is an infinite directed graph with one distinguished vertex $x_0$ and all other vertices connected with an edge to $x_0$ (and no other edges).
It is easy to see that $N$ is $\omega$-categorical
Now, consider $M$ which is the disjoint union of all finite connected graphs. It satisfies the existential part of theory of any graph (because any existential statement is witnessed by some finite graph) and it is countable (there are only countably many finite graphs, up to isomorphism), but it does not contain any infinite connected graph. In particular, it does not contain a copy of $N$.