Exercise 21.4 of Jech's Set Theory says:
Let $\kappa$ be an inaccessible cardinal. There is a notion of forcing $(P,<)$ such that $|P| = \kappa$ and $P$ is $\alpha$-distributive for all $\alpha < \kappa$, and such that $\kappa$ is not a Mahlo cardinal in the generic extension.
The hint below says:
Forcing conditions are sets $p \subseteq \kappa$ such that $|p \cap \gamma| < \gamma$ for every regular $\gamma \leq \kappa$; $p \leq q$ if and only if $p$ is an end-extension of $q$, i.e., if $q = p \cap \alpha$ for some $\alpha$. To show that for any $\alpha < \kappa$, $P$ does not add any new $\alpha$-sequence, observe that for every $p$ there is a $q \leq p$ such that $P_q = \{r \in P : r \leq q\}$ is $\alpha$-closed.
While it was not difficult proving that the $P$ in the hint is $\alpha$-distributive, I'm not sure how we have $\kappa$ is not Mahlo in $V[G]$. My guess is that $P$ adds a stationary subset of $\kappa$ that does not intersect the regular cardinals below $\kappa$, so I tried the most straightforward approach of taking $\bigcup G$. Yet it does not seem to be club.
Best Answer
In $V[G]$, $X := \bigcup G$ is an unbounded subset of $\kappa$. Moreover, by the distributivity claim, $\kappa$ is still regular.
Consider the club $\lim (X) = \{ \alpha : X \cap \alpha \text{ is unbounded in } \alpha \}$ (This is always a club). If $S$ is stationary, where $S$ is the set of regular cardinals, then $X \cap S \neq \emptyset$. Say $\lambda \in X \cap S$. Then $\lambda$ is a regular cardinal and $C \cap \lambda$ is unbounded in $\lambda$. $\lambda$ also was a regular cardinal in $V$ (the property of being regular is downwards absolute). Then there must have been a condition $p \in G$, with $X \cap \lambda \subseteq p$. This is a contradiction.