Assume you are given two infinite first-order structures over a countable language $\sigma$ , $M_0,$ and $M_1$ such that $M_0$ is a substructure of $M_1$ (denoted by $M_0 \subseteq M_1$), but not an elementary one (i.e., $M_0 \not\preceq M_1$), and both are models of a $\sigma$-theory $T$. Suppose I want to transfer the relationship of $M_0$ and $M_1$ to other cardinalities. How can one easily prove that there have to exist $\sigma$-models $N_0$ an $N_1$ of $T$, both of size continuum, such that $N_0 \subseteq N_1$, but $N_0 \not\preceq N_1$?
It seems like there are two cases for each of the models $M_0$ and $M_1$ to be considered, depending on their cardinalities.
If $M_0$ and $M_1$ are both countable, we might want to have the following idea work: define a theory $T_0$ in a language expanded by adding to $\sigma$ constants for all elements of $M_1$ (a fortiori, of $M_0$) and continuum new constans $C_\alpha: \alpha < 2^{\aleph_0}$:
$$ T_0 := T + ElDiag(M_0) + Diag(M_1) + \{c_\alpha \neq c_\beta: \alpha < \beta < 2^{\aleph_0}\}.$$ By compactness the theory has a model (repeating the reasoning in the proof of the weak form of the Upward Loewenheim-Skolem, basically) of size at least continuum, and actually of size exactly continuum, by the strong form of the Downward Loewenheim-Skolem. Call this model $N_0$. It is (up to isomorphism) an elementary extension of $M_0$ and an extension (necessarily, not an elementary one) of $M_1$.
Now define further a theory $T_1$ (in an obvious expanded language) as: $$T_1:=T + Diag(N_0) + ElDiag(B) + \{d_\alpha \neq d_\beta: \alpha < \beta < 2^{\aleph_0}\}.$$ If we can prove the theory is consistent than, by Downward Loewenheim-Skolem, it has a model of size continuum, $N_1$, and this model is an elementary extension of $M_1$, and an extension of $N_0$. Necessarily, this extension is not an elementary one, since there is a sentence $\varphi$ in the language of the model $M_0$, i.e., with parameters naming the elements of the universe of $M_0$, each of which is also an element of $M_1$, such that $M_0 \models \varphi$ (when $M_0$ is seen as a structure in the expanded language), but $M_1 \models \neg \varphi$, and therefore $N_1 \models \neg \varphi$ as well, whereas since $N_0 \models ElDiag(M_0)$, $N_0 \models \varphi$.
To prove that $T_1$ is consistent one would like perhaps to use compactness. Well, the problem is that we would most likely like to do something along the followign lines: suppose, for reductio, that there is a sentence $\psi(c)$ in $Diag(N_0)$ such that $ElDiag(M_1) \vdash \neg \psi(c)$. Necessarily, $c$ does not belong to the language of $N_0$, so actually, by the generalization of constants we have $ElDiag(M_1) \vdash \forall x \neg \psi(x)$. Now, normally we'd like to say something along the lines that then $T$ somehow excludes such a possibility or something, but the problem is that $\psi$ can still contain constants naming the elements of $N_0$, so $T$ does not have any authority over it.
Is there some easy way around it or there is some additional assumptions that needs to be added to make this statement go through?
(If the models in question are not countable, then I guess we need an additional step using a Downward Loewenheim-Skolem)
This uses two tricks: the easy one is identifying a "point of non-elementarity" by a new constant symbol (or symbols), and the more complicated one is looking at relativizations of formulas to unary predicates (see below if you're not familiar with this last point).
Concretely, since $M_0\not\preccurlyeq M_1$ there is some tuple $\overline{a}\in M_0$ and some $\Sigma$-formula $\varphi$ such that $M_0\models\varphi(\overline{a})$ but $M_1\models\neg\varphi(\overline{a})$. We then consider the complete theory $S$ of the expansion $N$ of $M_1$ by $(i)$ a new tuple of constant symbols $\overline{c}$ naming $\overline{a}$ and $(ii)$ a new unary relation symbol $U$ naming the underlying set of $M_0$ in $M_1$. This $S$ is a first-order theory in a countable language, so has a model $A\models S$ such that $\vert A\vert=\vert U^A\vert=2^{\aleph_0}$. Now let $N_1$ be the $\Sigma$-reduct of $A$, and let $N_0$ be the $\Sigma$-reduct of $U^A$. We get $N_0\subseteq N_1$ trivially, $N_0,N_1\models T$, and $N_0\not\preccurlyeq N_1$ - this last point because $\varphi^U(\overline{c})$ is in $S$. Note the use of relativization here: it's crucial that we use the formula $\varphi^U$ rather than $\varphi$ itself (since of course $\varphi(\overline{c})$ is not in $S$).
In case you haven't seen relativization before, the point is that $\theta^V$ is defined inductively for each formula $\theta$ and each unary predicate symbol $V$:
we set $\theta^V=\theta$ if $\theta$ is quantifier-free,
Booleans are boring (i.e. $(\psi\wedge\chi)^V\equiv\psi^V\wedge\chi^V$ and so on), and
$(\forall x\psi)^V\equiv\forall x(V(x)\rightarrow\psi)$ and $(\exists x\psi)^V\equiv \exists x(V(x)\wedge\psi)$ for each variable $x$.
The crucial fact about relativization, which is a good exercise, is the following:
EDIT: Note that I used Lowenheim-Skolem in a slightly different way than it's usually phrased above: I didn't just require $\vert A\vert=2^{\aleph_0}$, but simultaneously $\vert U^A\vert=2^{\aleph_0}$. This is of course an easy corollary of the theorem as usually stated (first get a model $B$ where $U^B=2^{\aleph_0}$, then take a submodel of $B$ with the same $U$ by adding constant symbols), but search for the term "two-cardinal theorems" and "Chang's conjecture" to see why simultaneous cardinality requirements are generally not to be glossed over (the saving grace here is that we're "keeping every set big" in a sense).