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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, if $G\le S_\omega$ is a maximal abelian subgroup, then $G=C(G)$: on the one hand, $G\le C(G)$ as $G$ is abelian. On the other hand, if $G\lneq C(G)$, we can fix $h\in C(G)\smallsetminus G$; then the subgroup generated by $G\cup\{h\}$ is abelian, contradicting the maximality of $G$. Thus, $|G|\le\aleph_0$ or $|G|=2^{\aleph_0}$ by the previous paragraph.

As noted in the comments by YCor, another way to set up the argument is to show that the closure of an abelian subgroup is abelian. This implies that a maximal abelian subgroup is closed.

This argument applies to any Polish group in place of $S_\omega$.

Here is a 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, if $G\le S_\omega$ is a maximal abelian subgroup, then $G=C(G)$: on the one hand, $G\le C(G)$ as $G$ is abelian. On the other hand, if $G\lneq C(G)$, we can fix $h\in C(G)\smallsetminus G$; then the subgroup generated by $G\cup\{h\}$ is abelian, contradicting the maximality of $G$. Thus, $|G|\le\aleph_0$ or $|G|=2^{\aleph_0}$ by the previous paragraph.

As noted in the comments by YCor, another way to set up the argument is to show that the closure of an abelian subgroup is abelian. This implies that a maximal abelian subgroup is closed.

This argument applies to any Polish group in place of $S_\omega$.

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, if $G\le S_\omega$ is a maximal abelian subgroup, then $G=C(G)$: on the one hand, $G\le C(G)$ as $G$ is abelian. On the other hand, if $G\lneq C(G)$, we can fix $h\in C(G)\smallsetminus G$; then the subgroup generated by $G\cup\{h\}$ is abelian, contradicting the maximality of $G$. Thus, $|G|\le\aleph_0$ or $|G|=2^{\aleph_0}$ by the previous paragraph.

This argument applies to any Polish group in place of $S_\omega$.

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, if $G\le S_\omega$ is a maximal abelian subgroup, then $G=C(G)$: on the one hand, $G\le C(G)$ as $G$ is abelian. On the other hand, if $G\lneq C(G)$, we can fix $h\in C(G)\smallsetminus G$; then the subgroup generated by $G\cup\{h\}$ is abelian, contradicting the maximality of $G$. Thus, $|G|\le\aleph_0$ or $|G|=2^{\aleph_0}$ by the previous paragraph.

As noted in the comments by YCor, another way to set up the argument is to show that the closure of an abelian subgroup is abelian. This implies that a maximal abelian subgroup is closed.

This argument applies to any Polish group in place of $S_\omega$.

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 $S_\omega$.

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, if $G\le S_\omega$ is a maximal abelian subgroup, then $G=C(G)$: on the one hand, $G\le C(G)$ as $G$ is abelian. On the other hand, if $G\lneq C(G)$, we can fix $h\in C(G)\smallsetminus G$; then the subgroup generated by $G\cup\{h\}$ is abelian, contradicting the maximality of $G$. Thus, $|G|\le\aleph_0$ or $|G|=2^{\aleph_0}$ by the previous paragraph.

This argument applies to any Polish group in place of $S_\omega$.