This question has been already answered thoroughly. I just wanted to address the OP's comment "I am not sure if there is anything special about $\aleph_\omega$".

Actually, there is nothing special about $\aleph_\omega$ other than the fact that it's a singular cardinal. Let $\kappa$ be a cardinal and let $S(\kappa)$ be the following statement:

>There is a family $\mathcal{F} \subset [\kappa]^{<\kappa}$ such that $|\mathcal{F}|=\kappa$ and for every $F \in [\kappa]^{<\kappa}$ there is $G \in \mathcal{F}$ such that $F \subset G$. 

Then $S(\kappa)$ holds if and only if $\kappa$ is a regular cardinal.

But things become more complicated if we just consider subsets of $\kappa$ of bounded cardinality smaller than $\kappa$. For example, let $C(\kappa)$ be the statement:

>There is a family $\mathcal{F} \subset [\kappa]^{\aleph_0}$ such that $|\mathcal{F}|=\kappa$ and for every $F \in [\kappa]^{\aleph_0}$ there is $G \in \mathcal{F}$ such that $F \subset G$. 

Then $C(\aleph_n)$ is true for every $0< n< \omega$, $C(\aleph_\omega)$ is false for essentially the same reason that $S(\aleph_\omega)$ is false, but the truth value of $C(\aleph_{\omega+1})$ depends on your set theory. Namely, if $\aleph_{\omega+1}$ countable subsets of $\aleph_\omega$ are enough to cover all countable subsets of $\aleph_\omega$ (for example, under the Generalized Continuum Hypothesis) then  $C(\aleph_{\omega+1})$ is true, while if at least $\aleph_{\omega+2}$ countable subsets of $\aleph_\omega$ are required to cover all countable subsets of $\aleph_\omega$ (which is possible, modulo large cardinals) then $C(\aleph_{\omega+1})$ is clearly false...