Linear suborders of $(P(\omega),\subseteq)$

Question

Consider the partial order $(P(\omega),\subseteq)$. Let $L$ be a dense linear suborder. Does $L$ have a countable dense subset?

(Note that it contains a copy of $\mathbb R$, via Dedekind cuts of $\mathbb Q$.)

Answer 1

For any $a\neq b\in\omega$ pick a set $A_{ab}\in L$ such that $a\in A_{ab},b\not\in A_{ab}$, if such exists, and let $D$ be the set of all $A_{ab}$ we have picked. We claim $D$ is dense in $L$.

Let $X,Y\in L$ with $X\subsetneq Y$. Pick any $Z$ such that $X\subsetneq Z\subsetneq Y$ and take $a\in Z\setminus X,b\in Y\setminus Z$. Then $Z$ is an element containing $a$ but not $b$, so $A_{ab}\in D$ exists. We have $A_{ab}\not\subseteq X$, so $X\subseteq A_{ab}$, and $Y\not\subseteq A_{ab}$, so $A_{ab}\subseteq Y$. Hence $D$ contains an element $A_{ab}$ strictly between $X$ and $Y$, so $D$ is a dense countable subset of $L$.

Note that countable choice was used in the proof, but it is unavoidable, since (\mathbb R and hence also) $P(\omega)$ can contain a finite Dedekind-infinite subset, and this will necessarily contain no dense countable subset.

Answer 2

Note that the lexicographic order of $2^\omega$ is a linear extension of $(\mathcal P(\omega),\subseteq)$, namely if $A\subseteq B$, then $\min(A\mathbin{\triangle}B)\in B$, which means that the first point that the sets differ must be in $B$.

But $2^\omega$ is hereditarily separable, and therefore every linear suborder of $\mathcal P(\omega)$ is separable.