Let $\kappa$ a cardinal of cofinality $\omega$; let $C \subseteq \kappa$ be a unbounded countable subset. Why is then $C$ closed (and thus a c.u.b.)? This means that if $\delta < \kappa$ is a limit ordinal, and $C \cap \delta$ is unbounded in $\delta$, then $\delta \in C$.
I doubt that this is true; however, it's a short statement in Kunen's set theory book.
On
If you set $\kappa = \aleph_{\omega}$ and let $C = \{0,,1,\aleph_1, 2, \aleph_2,...\}$ (so $C$ is the union of all natural numbers together with all $\aleph_n$ for $n>0$) and you let $\delta = \omega$, I think you have a counterexample.
$C\cap\delta = \delta$, so is automatically unbounded in $\delta$. But $\delta$ itself isn't in $C$ by construction.
What Kunen meant was that if $C$ is the range of a countable increasing unbounded sequence in $\kappa$, that is, $C$ is unbounded and has order type $\omega$, then it will be vacuously closed, since it has no limit points below $\kappa$.
This situation shows that the concept of club on cardinals of countable cofinality doesn't work as we want.