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.
Best Answer
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.
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.