Here is an alternative proof that no maximal abelian subgroup of $S_\omega$ can have cardinality strictly between $\aleph_0$ and $2^{\aleph_0}$.
Consider the standard product topology that makes $S_\omega$ a Polish group. For any subgroup $G\le S_\omega$, the centralizer $C(G)=\{h\in S_\omega:\forall g\in G\,gh=hg\}$ is closed, because if $h$ does not commute with some $g\in G$, this can be witnessed by finitely many values of $h$. This implies $|C(G)|\le\aleph_0$ or $|C(G)|=2^{\aleph_0}$ by the Cantor–Bendixson theorem.
Now, let $G\le S_\omega$ be a maximal abelian subgroup of cardinality $\aleph_0<\kappa<2^{\aleph_0}$. Then $G\le C(G)$, thus by the previous paragraph, $|C(G)|=2^{\aleph_0}$, and in particular, $G\lneq C(G)$. That is, there exists $h\in S_\omega\smallsetminus G$ that commutes with all elements of $G$, hence the subgroup generated by $G\cup\{h\}$ is abelian, contradicting the maximality of $G$.
This argument applies to any Polish group in place of $S_\omega$.