I'd like to prove that $\Vdash_{Coll(\omega_1, \kappa)} \Diamond$ where $\kappa$ is the cardinal $2^{\aleph_0}$
My attempt:
Recall that $Coll(\omega_1, \kappa)$ consists of functions $\omega_1 \rightarrow \kappa$ with domain of size $<\omega_1$.
Given an $M$-generic $G$ and corresponding $M[G]$, we first start with a surjection $g:\omega_1 \rightarrow \kappa$ where $g = \cup G \in M[G]$. We may also use basic cardinal arithmetic to find that, in $M[G]$, there is a bijection $h:\kappa \rightarrow [\omega_1]^{<\omega}$.
We define $A_{\alpha} = h(g(\alpha)) \cap \alpha$ for every $\alpha < \omega_1$. I'm trying to show that $\langle A_{\alpha}: \alpha < \omega_1 \rangle$ is our $\Diamond$ sequence. But I'm struggling to make this part work.
Would appreciate the help!
(There was an error in my previous answer: my original attempt is reducing our goal to $\mathsf{Add}(\omega_1,1)$ forces $\diamondsuit$, by showing every $V[G]$ for a $\operatorname{Col}(\omega_1,\kappa)$-generic filter contains a $\operatorname{Add}(\omega,1)$-generic filter $H$. The flaw is that there is no reason to believe $V[G]$ satisfies $\diamondsuit$ even if $V[H]$ does. However, a slight modification still works.)
Let me work over $V$ instead of $M$. Observe the following facts:
If $G$ is a $\operatorname{Col}(\omega_1,\kappa)$-generic filter over $V$, then $V[G]$ contains $\operatorname{Add}(\omega_1,1)$-generic filter over $V$. In fact, the function $\varphi: \operatorname{Col}(\omega_1,\kappa)=\operatorname{Fn}(\omega_1,\kappa,\omega_1)\to \operatorname{Add}(\omega_1,1)=\operatorname{Fn}(\omega_1,2,\omega_1)$ given by $\varphi(p)(\alpha) := p(\alpha)\mod 2$, that is, $$\varphi(p)(\alpha)=\begin{cases}0 & \text{if $p(\alpha)$ is of the form $\xi+2n$ for some $n\in\omega$,}\\ 1 & \text{otherwise.}\end{cases}$$ satisfies the following: if $D\subseteq\operatorname{Add}(\omega_1,1)$ is dense, then $\varphi^{-1}[D]$ is also a dense subset of $\operatorname{Col}(\omega_1,\kappa)$. Hence $\varphi^"[G]$ generates a generic filter of $\operatorname{Add}(\omega_1,1)$.
$\operatorname{Add}(\omega_1,1)$ forces $\diamondsuit$.
Hence we will mimic the proof of $\Vdash_{\operatorname{Add}(\omega_1,1)}\diamondsuit$. I borrowed the following proof from Kunen's new forcing textbook. Let us replace $\operatorname{Col}(\omega_1,\kappa)$ to $\operatorname{Col}(I,\kappa)$, where $$I=\{(\xi,\alpha)\mid \xi<\alpha<\omega_1\}.$$ For a generic filter $G$, let $f_\alpha(\xi):=(\bigcup G)(\xi,\alpha)\mod 2$, that is, $f_\alpha(\xi)=\varphi\circ (\bigcup G)(\alpha,\xi)$. Then $f_\alpha:\alpha\to 2$ is a characteristic function of some $A_\alpha\subseteq \alpha$. We claim that $\langle A_\alpha\mid\alpha<\omega_1\rangle$ is a $\diamondsuit$-sequence.
Work over $V[G]$, fix $h:\omega_1\to 2$ and a club $C\subseteq\omega_1$. Take a name $\dot{h}$, $\dot{C}$ and a condition $p_0$ such that $$p_0\Vdash \dot{h}:\omega_1\to 2\text{ and $\dot{C}$ is a club.}$$
Now we forget about $V[G]$ and work over $V$. For each condition $p$, let $\operatorname{supp}p$ be the least ordinal $\beta$ such that $\operatorname{dom}p\subseteq \{(\xi,\alpha)\mid \xi<\alpha<\beta\}$. Now define $p_n$, $\beta_n$, $\delta_n$, and $k_n$ such that
(Here we need that $\operatorname{Col}(\omega_1,\kappa)$ does not add any sequence of countable length, so for each ordinal $\beta$ and a condition $q$, we can find $r\le q$ and $k$ such that $r\Vdash \dot{h}\upharpoonright\beta=k$.)
Now $p=\bigcup_n p_n$, $\gamma=\bigcup_n\beta_n=\bigcup_n\delta_n$, and $k=\bigcup_n k_n$. Then $p$ forces $\gamma\in\dot{C}$ and $\dot{h}\upharpoonright\gamma=k$. Moreover, $\operatorname{supp}p=\gamma$. Now extend $p$ to $s$ by letting $$s(\xi,\gamma):= k(\xi).$$
Then $s\Vdash \dot{h}\upharpoonright\gamma=\dot{f}_\gamma$. Hence we proved the following:
Fix $\dot{h}$ and $\dot{C}$, the set of all such $s$ forms a dense set. Hence $V[G]$ thinks $\langle f_\alpha\mid\alpha<\omega_1\rangle$ is a $\diamondsuit$-sequence.