[proofplan] We use the defining reflection property of a Mahlo cardinal: the strongly inaccessible cardinals below $\kappa$ form a stationary subset of $\kappa$. A stationary subset of $\kappa$ meets every closed unbounded subset of $\kappa$. Applying this to the given club $C$ produces a cardinal $\delta$ lying both in $C$ and in the set of strongly inaccessible cardinals below $\kappa$. [/proofplan] [step:Form the stationary set of inaccessible cardinals below $\kappa$] Define \begin{align*} S := \{\delta \in \kappa : \delta \text{ is a strongly inaccessible cardinal}\}. \end{align*} Since $\kappa$ is Mahlo, the set $S$ is stationary in $\kappa$. [guided] We isolate the part of Mahloness that is relevant to the argument. Define the subset \begin{align*} S := \{\delta \in \kappa : \delta \text{ is a strongly inaccessible cardinal}\}. \end{align*} Here $S \subset \kappa$, and membership $\delta \in S$ means exactly that $\delta < \kappa$ and $\delta$ is strongly inaccessible. By the definition of a Mahlo cardinal, the collection of strongly inaccessible cardinals below $\kappa$ is stationary in $\kappa$. Therefore the set $S$ just defined is stationary in $\kappa$. This is the only point where the Mahloness hypothesis is used. [/guided] [/step] [step:Intersect the stationary set with the given club] The hypothesis says that $C \subset \kappa$ is closed and unbounded in $\kappa$, that is, $C$ is a club subset of $\kappa$. Since $S$ is stationary in $\kappa$, and stationary subsets of $\kappa$ meet every club subset of $\kappa$, we have \begin{align*} S \cap C \neq \varnothing. \end{align*} Choose $\delta \in S \cap C$. [/step] [step:Read off the required inaccessible cardinal] Because $\delta \in C$, the chosen ordinal lies in the given club. Because $\delta \in S$, the definition of $S$ gives that $\delta < \kappa$ and $\delta$ is a strongly inaccessible cardinal. Hence there exists a strongly inaccessible cardinal $\delta \in C$, as required. [/step]