Forcing square introduces diamond Let $\mathbb S_\kappa$ be the standard forcing for $\square_\kappa$ by initial segments.  This is $(\kappa+1)$-strategically closed.

Observation:  Let $T \subseteq \kappa^+$ be stationary.  If $T$ concentrates on $\mathrm{cof}({<}\kappa)$,then $\mathbb S_\kappa$ forces $\diamondsuit(T)$.  If $T$ concentrates on $\mathrm{cof}(\kappa)$ and $T$ is approachable, then $\mathbb S_\kappa$ forces $\diamondsuit(T)$.

The proof is like the argument that adding a Cohen subset of a regular cardinal $\kappa$ forces $\diamondsuit_\kappa$.  The coding goes by fixing bijections $f_\alpha : \alpha \to \kappa$ for $\alpha <\kappa^+$ and taking the diamond sequence to be $a_\alpha =  \{ \beta<\alpha: \alpha+f_\alpha(\beta)+1 \in C_{\alpha+\kappa} \}$.  It works because of the freedom we have in choosing $C_{\alpha+\kappa}$.  In the $\mathrm{cof}(\kappa)$-concentration case, we use an elementary submodel argument plus approachability so that we can work with a chain of conditions of length only $\kappa+1$.
Question 1: This seems like it might have been observed before. Does it appear in a paper?
Question 2: Can we eliminate the approachability assumption?
 A: Regarding Question 2, the assumption of approachability is necessary (or, more precisely, the assumption that $T$ has a stationary subset that is approachable), at least if we have $2^\kappa = \kappa^+$ in the ground model. The reason is that, if $T$ does not have a stationary subset that is approachable, then it becomes non-stationary after forcing with $\mathbb{S}_\kappa$, so $\diamondsuit(T)$ necessarily fails.
To see this, note that if $2^\kappa = \kappa^+$, then we can fix in $V$ an enumeration $\vec{a} = \langle a_\alpha \mid \alpha < \kappa^+ \rangle$ of all bounded subsets of $\kappa^+$. Then a set $S \subseteq \kappa^+$ is in $I[\kappa^+]$ if and only if there is a club $C \subseteq \kappa^+$ such that $\gamma$ is approachable with respect to $\vec{a}$ for every $\gamma \in S \cap C$. This is because, if $\vec{b} = \langle b_\alpha \mid \alpha < \kappa^+ \rangle$ is any other sequence of bounded subsets of $\kappa^+$, then there is a club of $\delta < \kappa^+$ such that all entries in $\langle b_\alpha \mid \alpha < \delta \rangle$ appear in $\langle a_\alpha \mid \alpha < \delta \rangle$. But now the assumption that $T$ does not have a stationary subset in $I[\kappa^+]$ means that there is in fact a club $C \subseteq \kappa^+$ such that, for every $\gamma \in T \cap C$, $\gamma$ is not approachable with respect to $\vec{a}$. Forcing with $\mathbb{S}_\kappa$ does not add any new bounded subsets to $\kappa^+$, so if $G$ is generic, then $\vec{a}$ remains an enumeration of all bounded subsets of $\kappa^+$ in $V[G]$ and every ordinal $\gamma < \kappa^+$ is approachable with respect to $\vec{a}$ in $V[G]$ if and only if it is approachable with respect to $\vec{a}$ in $V$. However, $\square_\kappa$ holds in $V[G]$, so also $\mathrm{AP}_\kappa$ holds. Then, in $V[G]$, $T \cap C \in I[\kappa^+]$, but every ordinal in this set is not approachable with respect to $\vec{a}$ so, there must be a club $D \subseteq \kappa^+$ disjoint from $T \cap C$. Then $C \cap D$ is disjoint from $T$, so $T$ is non-stationary.
Regarding Question 1, I made a similar but slightly different observation in my paper "Aronszajn trees, square principles, and stationary reflection" (see Lemma 3.11). I was forcing $\square(\lambda)$ rather than $\square_\kappa$, which slightly simplifies things. I just state the lemma for $\diamondsuit(\lambda)$, but the same argument works for $\diamondsuit(T)$ for any stationary $T \subseteq \lambda$. Approachability is not needed there, since the forcing notion for $\square(\lambda)$ is fully $\lambda$-strategically closed. I'm sure this had been recognized before, but I haven't seen it elsewhere in print.
