Let $\kappa$ regular infinite and $f: \kappa \rightarrow \kappa$ and take $$C_f :=\{\alpha < \kappa: f[\alpha]\subset\alpha\}$$
I want to show $C_f$ is closed unbounded. It's clarely closed: if $\alpha_1 < \cdots < \alpha_\eta < \cdots$ for $\eta < \delta < \kappa$ are in $C_f$ then $\alpha = sup \, \alpha_\eta$ is in $C_f$ because for all $\beta < \alpha$ there exist $\alpha_\eta > \beta$ thus $f(\beta)< \alpha_\eta < \alpha$. I had have some problems to show $C_f$ is unbounded.
This is not quite right. $\kappa = \omega$ is a counterexample. It's clear that, for, say, $f(n) = n+1,$ $C_f$ is not unbounded (in fact, it is empty in this case). However, for any regular cardinal $\kappa > \omega,$ it is unbounded.
To see this, let $\beta < \kappa$ and $\beta\notin C_f.$ We need to show there is an $\alpha$ with $\beta < \alpha < \kappa$ such that $\alpha\in C_f. $ Let $\beta_0=\beta.$ Then we can define $\beta_1 = \sup (f[\beta_0])+1.$ Then we have $\beta_0 < \beta_1 < \kappa.$ If $\beta_1\in C_f,$ we're done. Else, we define $\beta_2 = \sup(f[\beta_1])+1,$ and so on. If none of the $\beta_n$ are in $C_f,$ let $\alpha = \sup_{n\in\omega} \beta_n.$ So if $\gamma < \alpha,$ then $\gamma < \beta_n$ for some $n,$ so $f(\gamma) < \beta_{n+1} <\alpha.$ So we have $\beta < \alpha < \kappa,$ and $\alpha \in C_f.$
We used that $\kappa$ is regular here, and also that it is greater than $\omega$ and thus has cofinality greater than $\omega$... I'll let you spot where.