Assume ZFC.
I read a claim:
"[A]ny singular cardinal contains a closed unbounded set consisting of ordinals that are not cardinals, [...]"
and I wanted to prove that claim. Here is the argument I came up with:
Say $\aleph_\alpha$ is a singular cardinal so $\alpha$ is a limit. For each $\eta<\alpha$, let $C_\eta$ be closed unbounded in $\aleph_\eta$ and we may assume $0\notin C_\eta$. Consider $E_\eta=\{ \aleph_\eta+\beta\mid \beta\in C_\eta\}$; then $E_\eta\subset(\aleph_\eta,\aleph_{\eta+1})$ for each $\eta$. If we consider $E=\bigcup_{\eta<\alpha}E_\eta$, then $E$ is closed unbounded in $\aleph_\alpha$: Clearly it is unbounded. Suppose $\gamma<\aleph_\alpha$ and $\gamma\cap E$ is unbounded in $\gamma$. Since there is some $\eta$ such that $\gamma<\aleph_\eta$, if $\eta$ is the least such it must have a predecessor $\xi$, and so $\aleph_\xi<\gamma<\aleph_{\xi+1}$. Then we know $\gamma\cap E_\xi$ is unbounded in $\gamma$ and so $\gamma\in E_\xi$.
This argument should be wrong because I didn't use the assumption that $\aleph_\alpha$ is singular, and Mahlo cardinal (a strongly inaccessible cardinal $\kappa$ such that the set of regular cardinals under $\kappa$ is stationary in $\kappa$) is a thing. So where is the problem?