I'm sure this is a silly question. Suppose $\lambda$ is a singular cardinal of uncountable cofinality. Then surely $\diamondsuit_\lambda$ must fail. But why?
Just in case, let me specify that by $\diamondsuit_\lambda$ I mean that there is a sequence of sets $A_\alpha\subseteq \alpha$ for $\alpha<\lambda$ such that for any $A\subseteq\lambda$ the set $\{\alpha<\lambda;A\cap\alpha=A_\alpha\}$ is stationary in $\lambda$.
I vaguely remember thinking about this some time ago and convincing myself that $\diamondsuit_\lambda$ should violate König's theorem. But I can't reconstruct the argument. Certainly, $\diamondsuit_\lambda$ implies that there are only $\lambda$ many bounded subsets of $\lambda$. If $\lambda$ was regular then this would also mean that $\lambda^{<\lambda}=\lambda$, since any short sequence in $\lambda$ is bounded. But if $\lambda$ is singular then there are elements of ${}^{<\lambda}\lambda$ which are unbounded in $\lambda$, so the previous trick doesn't apply.
Assume diamond holds at $\lambda$. Since $\lambda$ has a club of size $cf(\lambda)$, there cannot be more than $2^{cf(\lambda)} \leq \lambda$ many almost disjoint stationary subsets of $\lambda$. There are at least $\lambda^+$ many different subsets of $\lambda$, each guessed on some stationary subset of $\lambda$ and these sets are almost disjoint - A contradiction.