Let $\kappa$ be a regular cardinal, equipped with order topology, and $C\subseteq\kappa$ a closed unbounded subset.
If $\alpha<\kappa$ is another regular cardinal and a limit point of $C$, is the set $C\cap\alpha$ closed and unbounded again? In Jech, Set Theory, pag.95, it seems that this is due to the regularity of $\alpha$ (clearly, the unboundness follows from $\alpha$ being a limit point of $C$), but my question is if this obviously follows from the definition of relative topology on $\alpha\subseteq\kappa$.
How about the set $C'\cap\alpha$ (where $C'$ is the derived set of $C$, i.e. the set of all limit points of $C$)? Is it closed for the same reason?
For the first question, you're right: if $C$ is closed in $\kappa$ and $C\cap\alpha$ is unbounded in $\alpha$, then $C\cap\alpha$ is closed in $\alpha$.
For the second question, however, an additional assumption is indeed needed. For example, suppose $C$ is a club in $\omega_{\omega+1}$ (which is regular) with $C\cap \omega_\omega=\{\omega_i: i\in\omega\}$. Then $C'\cap\omega_\omega=\emptyset$.
Regularity (and uncountability) solves this problem, but is in fact not necessary: if $cf(\alpha)$ is uncountable and $C$ is a club in $\kappa$ which is unbounded in $\alpha$, then $C'\cap\alpha$ is again unbounded in $\alpha$. However, when we iterate the derivative we see a problem: $\alpha^{(cf(\alpha))}\cap \alpha$ is empty for all $\alpha$.
In fact, the following are equivalent:
For all $C\subseteq\alpha$ club and all $\beta<\alpha$, $C^{(\beta)}$ is club in $\alpha$.
$\alpha$ is uncountable and regular.
Here "$X^{(\gamma)}$" denotes the $\gamma$th derivative of $X$, defined recursively as $X^{(0)}=X$, $X^{(\gamma+1)}=(X^{(\gamma)})'$, and $X^{(\lambda)}=\bigcap_{\gamma<\lambda}X^{(\gamma)}$ for limit $\lambda>0$.