Let $R$ be a countable set and $T$ the set of functions $R \to \{0,1\}$ with the pointwise order. ($T$ is chain-complete, that is, every chain of $T$ has a least upper bound - the pointwise maximum.)
I suspect that every chain of $T$ is countable, and I have come up with an informal proof, which however seems to have some issues, and which I have trouble formalizing:
Let $C$ be an uncountable chain of $T$. For any distinct $c_0,c_1 \in C$, if $c_0 \leq c_1$, $c_1$ must have strictly more elements $r$ such that $c_1(r)=1$. Therefore, the least upper bound of $C$ must have uncountably many such elements. However, $R$ is countable.
Question: Can this be made more formal, or is it incorrect?
One way I tried to make the "proof" into an actual proof was to try to build a bijection between elements of $C$ and some subset of $R$ like this: map $c$ to an element of $R$ which maps to 1 under $c$ but not under its predecessor. However, this seems to fail for limit elements of $C$. (I suspect $C$ would be uncountable even with those limit elements removed, but I am not sure about that at all.)
I have also thought of using some kind of induction, however my problem is that the order on $T$ need not be well-founded, eg. take $R = \mathbb{N}$ and the chain $$\{n \mapsto n \geq m \mid m \in \mathbb{N} \}\,.$$
(Given a chain $C$, a predecessor of $c \in C$ is the largest element of $C$ less than $c$. If such an element does not exist, $c$ is a limit element.)
This is in fact not true. For instance, let $R=\mathbb{Q}$. For each $r\in\mathbb{R}$, let $f_r:\mathbb{Q}\to\{0,1\}$ be the function defined by $f_r(x)=0$ if $x>r$ and $f_r(x)=1$ if $x\leq r$. Then $f_r\leq f_s$ pointwise iff $r\leq s$, so $\{f_r:r\in\mathbb{R}\}\subset T$ is a chain which is uncountable (and order-isomorphic to $\mathbb{R}$).
What your idea shows is instead that any chain $C\subseteq T$ has what is (somewhat confusingly) known as the countable chain condition. This means that any collection of pairwise disjoint open intervals $(a_i,b_i)$ in $C$ is countable (since for each $i$ you can pick $r_i\in R$ such that $a_i(r_i)=0$ and $b_i(r_i)=1$, and these $r_i$ must all be distinct). However, an uncountable totally ordered set can still have the countable chain condition (as exemplified by $\mathbb{R}$).