Is it true the statement that: if $\kappa$ is a Mahlo cardinal, then for all $\alpha<\kappa$ there is a cardinal $\lambda$ with $\alpha<\lambda<\kappa$ and $\lambda$ contains $\lambda$-many inaccessible cardinals?
It seems really obvious to me, but I tried to prove it and it feels a bit odd.
Yes, this is true. Here's one way to show this.
Lemma. If $\lambda$ is inaccessible, then $\lambda$ is the $\lambda$th inaccessible if and only if the set of inaccessible cardinals below $\lambda$ is unbounded if and only if $V_\lambda\models$"There is a proper class of inaccessible cardinals".
This lemma is easy enough to prove, so I will leave you to work it out.
Exercise. If $\kappa$ is Mahlo, then $\kappa$ is the $\kappa$th inaccessible cardinal.
Now, show that $C=\{\alpha<\kappa\mid (V_\alpha,\in)\prec (V_\kappa,\in)\}$ is a club, where $\prec$ means "elementary submodel" in the language of set theory. And now that "There is a proper class of inaccessible cardinals" is a first-order statement in the language of set theory.
Finally, since $\kappa$ is Mahlo, the inaccessible cardinals below it form a stationary set, so $\{\lambda\in C\mid\lambda\text{ is inaccessible}\}$ is a stationary set as well as the intersection of a club and a stationary set. In particular, it is unbounded. Now, apply the lemma.