Is $Th(\mathbb{Z}, <) \cup Diag (\mathbb{Q}, <)$ satisfiable? I want to show that there exist $\mathcal{M} \equiv (\mathbb{Z}, <)$ and an order-preserving embedding $\sigma :\mathbb{Q} \rightarrow \mathcal{M}$. 
It's enough to see that $ Th(\mathbb{Z}, <) \cup Diag (\mathbb{Q}, <)$ satisfiable and by compactness that each subset $\Sigma_0$ is satisfiable. WLOG, $\Sigma_0$ isthe single formula 
$$\varphi \wedge \psi(a_1, \cdots a_n)$$
such that $(\mathbb{Z}, <) \vDash \varphi$, $(\mathbb{Q}, <) \vDash\psi(a_1, \cdots a_n)$ and $\psi(a_1, \cdots a_n)$ is Boolean combination  of atomic formulas. Finally I know the atomic formulas are $x = y$ of $x < y$.
My question is: Is $(\mathbb{Z}, <)$ a model of $ Th(\mathbb{Z}, <) \cup Diag (\mathbb{Q}, <)$? Why?
 A: It doesn't make sense to talk about $(\mathbb Z, <)$ being a model of $\operatorname{Th}(\mathbb Z,<)\cup \operatorname{Diag}(\mathbb Q, <)$ without giving an interpretation of all the constant symbols (i.e. all the rationals) in $\mathbb Z$. If we could do this, and the result were a model, it would amount to finding order preserving embedding of $\mathbb Q$ into $\mathbb Z$ itself. But it's pretty clear that no such embedding exists.
So $\mathbb Q$ doesn't embed into just any model of $\operatorname{Th}(\mathbb Z, <).$ However, the problem is correct that there is some model that it embeds into, and since we know what all models of $\operatorname{Th}(\mathbb Z,<)$ look like, we can find an explicit example. The models of $\operatorname{Th}(\mathbb Z,<)$ are anything of the form $(L\times \mathbb Z, \prec),$ where $L$ is a linearly ordered set and $\prec$ is the dictionary order on $L\times \mathbb Z.$ So, for instance, $\mathbb Q\times \mathbb Z$ is a model of $\operatorname{Th}(\mathbb Z, <),$ and you can embed $\mathbb Q$ into that easily.
