$\newcommand{\c}[1]{\left\langle #1 \right\rangle}$ $\newcommand{\A}{\mathcal{A}}$ $\newcommand{\Po}{\mathcal{P}}$ I refer to Kunen's Set Theory, 2011 Edition, Theorem III.7.14.
Assume $V = L$. Then $\Diamond^+$, and hence $\Diamond$, is also true.
Below is a partial excerpt of Kunen's proof, slightly reworded for my own readability:
We first define the $\Diamond^+$-sequence, $\c{\A_\alpha : \alpha < \omega_1}$, as follows: For each $\alpha < \omega_1$, let: \begin{align*} \alpha^* := \min\{\delta : \alpha < \delta < \omega_1 \wedge L_\delta \preceq L_{\omega_1}\} \end{align*} that is, $\alpha^*$ is the smallest layer above $L_\alpha$ that is an elementary submodel of $L_{\omega_1}$. This exists due to Löwenheim-Skolem, which gives a countable elementary submodel containing $\alpha + 1$, and the condensation lemma, which gives that any elementary submodel must be $L_\delta$ of some $\delta < \omega_1$ (note that any elementary submodel of $L_{\omega_1}$ is transitive). We can now define the sequence by: $$ \A_\alpha := L_{\alpha^*} \cap \Po(\alpha) $$ Clearly $\A_\alpha$ is a countable subset of $\Po(\alpha)$. We now prove that $\A_\alpha$ gives a $\Diamond^+$-sequence. Fix $A \subseteq \omega_1$. Then $A \in L_{\omega_2}$ by $V = L$. Now for any $\sigma < \omega_1$, let $M_{A,\sigma} = \mathfrak{H}(L_{\omega_2}, \{A\} \cup \sigma)$ be the Skolem hull of $\{A\} \cup \sigma$ in $L_{\omega_2}$. Then $M_{A,\sigma} \preceq L_{\omega_2}$ as $L_{\omega_2}$ has a definable well-order (over the empty set). $M_{A,\sigma}$ has the properties: 1) $M_{A,\sigma}$ is countable, 2) $M_{A,\sigma} \cap \omega_1$ is a countable limit ordinal (by Lemma III.7.12(2)), and 3) $M_{A,\sigma} \cap \omega_1 \geq \sigma$. Finally, let: $$ C := C_A := \{\sigma < \omega_1 : M_{A,\sigma} \cap \omega_1 = \sigma\} $$ ...
Kunen then left showing that $C$ is a club as an exercise to the reader. However, I don't see any straightforward way to prove it. My main issue lies in not knowing how the Skolem hulls $M_{A,\sigma}$ look like, and there don't seem to be any result mentioned prior to this theorem that is useful either.
Any help is appreciated.
To see that $C$ is a club, observe that for limit ordinals $\alpha$: $$ M_{A,\alpha} = \bigcup_{\beta < \alpha} M_{A,\beta} $$ Then:
It is certainly closed, as if $M_{A,\alpha_n} \cap \omega_1 = \alpha_n$, and $\alpha = \sup{\alpha_n}$, then: $$ \alpha = \bigcup_{n \in \omega} \alpha_n = \bigcup_{n \in \omega} M_{A,\alpha_n} \cap \omega_1 = M_{A,\alpha} \cap \omega_1 $$
To see that it is unbounded, let $\alpha < \omega_1$, and let $\sigma_0 > \alpha$ be such that $M_{A,\sigma_0} \cap \omega_1 > \sigma_0$ (if no such $\sigma_0$ exists, then unboundedness follows immediately). Now define recursively $\sigma_{n+1} := M_{A,\sigma_n} \cap \omega_1$. If the sequence becomes constant eventually, then we are done. Otherwise, consider $\sigma = \sup_n \sigma_n$. Then $\sigma$ is a limit ordinal, and we have that $M_{A,\sigma} \cap \omega_1 = \sigma$.