I think there's a simpler way to answer
EDIT: As Andreas Blass points out below, this questionapproach doesn't work.
Suppose $\mathcal{F}=\lbrace X_\alpha: \alpha\in\aleph_\omega\rbrace$ were such a collection of sets. Now for $\alpha\in\aleph_\omega$, let $rank(\alpha)=\min\lbrace \beta: \alpha\in X_\beta\rbrace$, and let $\le_W\subseteq\aleph_\omega\times\aleph_\omega$ be a well-ordering of $\aleph_\omega$ with the property that $rank(\alpha)$ < $rank(\beta)\implies \alpha\le_W\beta$. Note that no $X_\alpha$ is cofinal in $\le_W$: since each $X_\alpha$ has size $<\aleph_\omega$, for each $\alpha$ the set $\bigcup_\beta\le\alpha X_\beta$ has size $<\aleph_\omega$, and hence (by the third assumption on $\mathcal{F}$) there is some $\gamma>\alpha$ with $X_\gamma\not\subseteq X_\alpha$; any element of $X_\gamma-X_\alpha$ is then $\le_W$-above each element of $X_\alpha$. Let $C\subseteq \aleph_\omega$ be countable and cofinal in the $\le_W$ ordering. Then clearly $C\not\subseteq X_\alpha$ for any $\alpha\in\aleph_\omega$ since no $X_\alpha$ is cofinal in $\le_W$. But $C$ has cardinality $\aleph_0<\aleph_\omega$; a contradiction.
To me this makes the role of singularity clear(er).
