Find an example of an constructible model. I have found the rigorous definition of constructible model like this:
Let $L$ be a first-order language and $M$ an $L$-structure. If $A\subseteq M$, $M$ is constructible over $A$ if $(M-A)$ can be written as $(c_i\colon i<\lambda)$, where $\lambda$ is an ordinal, such that, for any $j<\lambda$, the type $tp_M(c_j/A\cup \{b_i\mid i<\lambda\})$ is isolated.
I would find an example of constructible structure based on this.
 A: If the language $L$ is countable, then a structure $M$ is constructible (over $\varnothing$) if and only if it is countable and atomic if and only if it is a prime model of its complete theory $T = \mathrm{Th}(M)$. In fact, if $M$ is countable and atomic, then any enumeration $M = (a_n)_{n\in \omega}$ witnesses that $M$ is constructible. Why? For every $n$, since $M$ is atomic, $\text{tp}(a_0,\dots,a_n/\varnothing)$ is isolated by a formula $\varphi(x_0,\dots,x_n)$. Then $\text{tp}(a_n/a_0,\dots,a_{n-1})$ is isolated by the formula $\varphi(a_1,\dots,a_{n-1},x_n)$. Thus $M$ is constructible.
So examples abound over $\varnothing$ when the language is countable: Just pick any theory that has a prime model, equivalently a countable atomic model, and then enumerate this model by $\omega$ in any way you like. Recall that $T$ has a countable atomic model if and only if the isolated types are dense in $S_n(T)$ for all $n$.
What about constructible structures over non-empty sets? Well, if $A$ is any set, then a constructible $L$-structure over $A$ is the same thing as a constructible $L_A$-structure over $\varnothing$, where $L_A$ is the language $L$ extended by a new constant symbol naming every element of $A$. As long as $A$ is countable, the language stays countable, and the above discussion applies.
When the language is uncountable, or when we want to find constructible models over uncountable sets, the easiest way to find examples is by looking at totally transcendental theories (a theory is totally transcendental if every formula has ordinal-valued Morley rank - this is equivalent to $\omega$-stability when the language is countable). The theorem is that if $T$ is totally transcendental, and $A$ is any subset of a model $M\models T$, then $M$ has an elementary substructure that is constructible over $A$. Since a constructible model over $A$ is prime over $A$, and constructible models are unique up to isomorphism, it follows that any subset of a model of $T$ is contained in a canonical "smallest" model of $T$.
The proof tells you how to find the construction sequence. The key fact is that if $T$ is totally transcendental, then isolated types are dense in $S_n(A)$ for all sets $A$. Start with your set $A\subseteq M$. If it is not an elementary substructure of $M$, then by the Tarski-Vaught test, there is some formula $\varphi(x,a)$ with parameters $a$ from $A$ which is consistent but has no solution in $M$. Since isolated types are dense in $S_1(A)$, $\varphi(x,a)$ is contained in some type $p(x)\in S_1(A)$ which is isolated over $A$ and hence realized by some $b_0\in M$. Now repeat the argument over $Ab_0$. By transfinite induction, we construct an ordinal-indexed construction sequence $(b_\alpha)$, stopping when we reach an elementary substructure of $M$ which is constructible over $A$.
For an explicit example, let $T$ be the theory of an equivalence relation with infinitely many infinite classes (and no finite classes). This theory is totally transcendental. Let $A = (a_{\alpha})_{\alpha\in \aleph_1}$ be an uncountable set of pairwise non-equivalent elements. The constructible model over $A$ has one countably infinite class containing $a$ for each $a\in A$ and no other classes. For a construction sequence $(b_\beta)_{\beta\in \aleph_1}$, enumerate each countably infinite class in order-type $\omega$, so $(b_{\omega\cdot \alpha+n})_{n\in \omega}$ is an enumeration of the class of $a_{\alpha}$. The point is that the type of $b_{\omega\cdot \alpha+n}$ over the previous elements in the sequence is isolated by the formula $xEa_{\alpha}\land x\neq a_{\alpha}\land \bigwedge_{i=0}^{n-1}x\neq b_{\omega\cdot \alpha+k}$.
