Construct an elementary chain of order type $\mathbb Q$ Let $T$ be a complete first-order theory over a countable language $L$, such that $T$ has an infinite model.
It is well-known that we can construct an elementary chain of size-$\aleph_0$ models of order type the reverse ordering of any countable ordinal, see e.g. here. Simply, a possible proof uses Gödel's compactness theorem as well as augmenting the language with suitable Skolem functions.
It is even more well-known that we can construct an elementary chain of size-$\aleph_0$ models of order type any countable ordinal. Moreover, the ordinary or reversed ordering of any countable ordinal continuously embeds into the ordering of $\mathbb Q$.
Question 1: Is it always possible to construct an elementary chain $\{M_q:q\in\mathbb Q\}$ of models of $T$, such that every $\|M_q\|=\aleph_0$, every $M_p\precneqq M_q$ whenever $p<q$, and moreover for all $q\in\mathbb Q$, it holds that
\begin{equation}\tag{$\dagger$}\bigcup_{i<q}M_i=M_q=\bigcap_{j>q}M_j
\end{equation}
Question 2: If the answer to Question 1 is NO, then under what conditions for $T$ can the answer become YES?
Thanks in advance!

Simple observations:
(1) Every discontinuity of the chain must happen at Dedekind cuts of irrarional numbers. For example, if $T$ is the theory of infinite sets, then let $S$ be the set of all irrational algebraics, and the models $M_q=\{s\in S:s<q\}$ work as desired.
(2) By repeating the argument mentioned above, it is clear that the answer is YES when $(\dagger)$ is weakened to
$$\bigcup_{i<q}M_i\subseteq M_q\subseteq \bigcap_{j>q} M_j
$$
and it's easy to find examples such that both subset relations modified as above are cofinitely strict. Thus by replacing each $M_q$ with their upper/lower bounds, we see that the answer is also YES when either of the two equalities in $(\dagger)$ is omitted.
(3) It suffices for $(\dagger)$ to hold only for a dense subset of an interval in $\mathbb Q$, since $\mathbb Q$ would continuously embed into it.
 A: Here's a quick sketch (assuming WLOG that $T$ has definable Skolem functions):
Let $\mathbb{A}$ be the set of rational multiples of $\pi$. Suppose $\mathcal{N}$ is a "big" model of $T$. Pick some $\mathbb{A}$-indexed "very indiscernible" sequence $X=\{x_i: i\in\mathbb{A}\}$ of elements of $\mathcal{N}$. For each $q\in\mathbb{Q}$ we can then set $M_q$ to be the definable closure of $\{x_i: i\in\mathbb{A}, i<q\}$ in $\mathcal{N}$.
Clearly we have $p<q\implies M_p\preccurlyeq M_q$ and $M_q=\bigcup_{p<q}M_p=\bigcap_{p>q}M_p$ (this last bit is where $\mathbb{A}\cap\mathbb{Q}=\emptyset$ comes in), so we just have to ensure that $M_p\subsetneq M_q$. This will come down to the appropriate choice of $\mathcal{N}$ and $X$. Choosing $\mathcal{N}$ is easy since all we want is sufficient "richness:" we'll just pick any sufficiently saturated model of $T$.
Now we just have to pick an appropriate $X$. Here we run into the issue Alex Kruckman highlighted in the linked question: it's not the case that any old $\mathbb{A}$-indexed indiscernible sequence in $\mathcal{N}$ will do. Instead, we need something like a "coheir sequence of ordertype $\mathbb{Q}$." Here we can use a cute trick: if I have a structure $\mathcal{S}$ and a sequence $\sigma=(s_i)_{i\in\omega}$ of elements of $\mathcal{S}$, I can get a richer sequence similar to $\sigma$ by taking an ultrapower of $\mathcal{S}$ with $\sigma$. Precisely, let $\mathcal{S}_\sigma$ be the expansion of $\mathcal{S}$ by adding a unary predicate $U$ naming the elements in $\sigma$ and a binary relation $\triangleleft$ describing the ordering of those elements according to $\sigma$, and consider an appropriate ultrapower $\mathcal{U}$ of $\mathcal{S}_\sigma$ (or really just any sufficiently saturated model of $Th(\mathcal{S}_\sigma)$). The "expanded sequence" $(U^\mathcal{U},\triangleleft^\mathcal{U})$ now has very rich ordertype, and in particular has a suborder of type $\mathbb{A}$.
Here we can use this trick as follows: "blow up" a coheir sequence in $\mathcal{N}$ to get a $\mathbb{A}$-indexed sequence $X$ in some larger $\mathcal{N}'$ which we can check works for the same reason that a coheir sequence helped solve the problem in the linked question. OK fine, we're now using $\mathcal{N}'$ instead of $\mathcal{N}$ after all, but that's okay.
A: We can do much better! I'll explain how to construct an $\mathbb{R}\setminus \mathbb{Q}$-indexed chain of countable models with the continuity properties you want. This is more general, since you can get a $\mathbb{Q}$-indexed chain by taking the subchain indexed by $(\pi + \mathbb{Q})\subseteq \mathbb{R}\setminus \mathbb{Q}$.
$T$ is a complete first-order theory over a countable language $L$, with an infinite model. Let $T'$ be a Skolemization of $T$ (and note that the Skolemized language $L'$ is still countable). We'll work in $T'$, and then you can take the reduct of everything back to $L$ at the end of the day.
Let $I = (a_q)_{q\in \mathbb{Q}}$ be any non-constant indiscernible sequence in a model of $T'$. For $r\in \mathbb{R}\setminus \mathbb{Q}$, let $I_{<r} = \{a_q\mid q\in \mathbb{Q}, q<r\}$, and define $M_r = \text{dcl}(I_{<r})$. Now let's check your conditions.

*

*For any $r\in \mathbb{R}\setminus \mathbb{Q}$, $|M_r| = \aleph_0$, since $I_{<r}$ is countable and the definable closure (in a countable language) of a countable set is countable.

*When $r<s$ in $\mathbb{R}\setminus \mathbb{Q}$, $I_{<r} \subseteq I_{<s}$, so $M_r\subseteq M_s$, and since we have Skolem functions, $M_r\preceq M_s$. Now pick some $q\in\mathbb{Q}$ with $r<q<s$. Then $a_q\in I_{<s}\subseteq M_s$. To show that $M_r\precneqq M_s$, I claim that $a_q\notin M_r$. So suppose for contradiction that $a_q\in M_r = \text{dcl}(I_{<r})$. Then there are $a_{p_1},\dots,a_{p_k}\in I_{<r}$ and a formula $\varphi(a_{p_1},\dots,a_{p_k},y)$ defining $a_q$. Pick some $q'>q$ in $\mathbb{Q}$. By indiscernibility, we also have  $\varphi(a_{p_1},\dots,a_{p_k},a_{q'})$, so $a_q = a_{q'}$. But then by indiscernibility, the sequence $I$ is constant, which is a contradiction.

*Fix $r\in \mathbb{R}\setminus \mathbb{Q}$. We have $M_s\subseteq M_r$ for all  $s<r$, so $\bigcup_{s<r}M_s \subseteq M_r$. For the other inclusion, let $b\in M_r = \text{dcl}(I_{<r})$. Then there are $a_{p_1},\dots,a_{p_k}\in I_{<r}$ with $p_1<\dots<p_k<r$ and a formula $\varphi(a_{p_1},\dots,a_{p_k},y)$ defining $b$. Pick some $s\in \mathbb{R}\setminus \mathbb{Q}$ with $p_k < s < r$. Then $a_{p_1},\dots,a_{p_k}\in I_{<s}$, so $b\in \text{dcl}(I_{<s}) = M_s$. Thus $M_r\subseteq \bigcup_{s<r} M_s$.

*Again, fix $r\in \mathbb{R}\setminus \mathbb{Q}$. We have $M_r\subseteq M_s$ for all $s>r$, so $M_r\subseteq \bigcap_{s>r} M_s$.  For the other inclusion,  let $b\in \bigcap_{s>r} M_s$. Pick some $s>r$ in $\mathbb{R}\setminus \mathbb{Q}$, so $b\in M_s = \text{dcl}(I_{<s})$. Then there are $a_{p_1},\dots,a_{p_k},a_{q_1},\dots,a_{q_\ell}\in I_{<s}$ with $p_1<\dots<p_k<r<q_1<\dots<q_\ell<s$ and a formula $\varphi(a_{p_1},\dots,a_{p_k},a_{q_1},\dots,a_{q_\ell},y)$ defining $b$. Note that we use the fact that $r$ is irrational here: it is important that none of the $p_i$ or $q_j$ are equal to $r$. Pick some  $s'\in \mathbb{R}\setminus \mathbb{Q}$ with $r < s' < q_1$. Then also $b\in M_{s'} = \text{dcl}(I_{<s'})$, so there are $a_{p'_1},\dots,a_{p'_{k'}},a_{q'_1},\dots,a_{q'_{\ell'}}\in I_{<s'}$ with $p'_1<\dots<p'_{k'}<r<q'_1<\dots<q'_{\ell'}<s'$ and a formula $\psi(a_{p'_1},\dots,a_{p'_{k'}},a_{q'_1},\dots,a_{q'_{\ell'}},y)$ defining $b$. Then we have: $$\forall x\, (\varphi(a_{p_1},\dots,a_{p_k},a_{q_1},\dots,a_{q_\ell},y) \leftrightarrow \psi(a_{p'_1},\dots,a_{p'_{k'}},a_{q'_1},\dots,a_{q'_{\ell'}},y)).$$
Now we will slide the $a_{q'_i}$ into $M_r$. Let $p^* = \max(p_k,p'_{k'})<r$, and pick $q''_1,\dots,q''_{\ell'}\in \mathbb{Q}$ with $p^*<q''_1<\dots<q''_{\ell'} < r$. By indiscernibility, we have: $$\forall x\, (\varphi(a_{p_1},\dots,a_{p_k},a_{q_1},\dots,a_{q_\ell},y) \leftrightarrow \psi(a_{p'_1},\dots,a_{p'_{k'}},a_{q''_1},\dots,a_{q''_{\ell'}},y)).$$
In other words, $\psi(a_{p'_1},\dots,a_{p'_{k'}},a_{q''_1},\dots,a_{q''_{\ell'}},y)$ defines $b$. But $a_{p'_1},\dots,a_{p'_{k'}},a_{q''_1},\dots,a_{q''_{\ell'}}\in I_{<r}$, so  $b\in \text{dcl}(I_{<r}) = M_r$. Thus $\bigcap_{s>r} M_s\subseteq M_r$.


Note that the phrase "coheir sequence" didn't appear in this construction. In fact, the only thing special about our indiscernible sequence $I$ is that it was non-constant. The point is that the type of  any indiscernible sequence is finitely satisfiable in any initial segment without an upper bound,  just by indiscernibility. So the tail segment $I_{> r}$ is automatically a coheir sequence over $M_r = \text{dcl}(I_{<r})$.
The reason we had to pick our indiscernible sequence more carefully in the question you linked to is that there  we wanted to make an arbitrary model $M$ the intersection of an elementary chain of models. So we had to take special care to ensure that we built the elementary chain from a coheir sequence - the finite satisfiability in $M$ was not automatic the way it is for models that come from the indiscernible sequence itself.
