I am looking for a hint (not a solution) to exercise IV.7.28 of Kunen's Set Theory book (2013). Recall that a poset $\mathbb{P}$ is separative if for every $p,q\in \mathbb{P}$, $p\nleq q$ implies $\exists s\in \mathbb{P}$ such that $s\leq p$ and $s\perp q$. The exercise is as follows:
Let $M$ be a ctm for ZFC, let $\mathbb{P}$ be a poset in $M$ and $\theta$ a cardinal in $M$. Prove that (1)$\implies$(2) and (2)$\implies$(1) if $P$ is separative:
(1) In $\mathbb{P}$, every intersection of $\theta$ dense open sets is dense in $\mathbb{P}$.
(2) $\mathbb{P}$ doesn't add $\theta$-sequences.
The exercise asks the reader to interpret these statements in the ctm approach. The way I see it is that $\mathbb{P}$ doesn't add $\theta$-sequences if for any $G$ that is $\mathbb{P}$-generic over $M$ and every $f\in M[G]$ with dom($f$)=$\theta$ and ran($f$)$\subseteq M$ it holds that $f\in M$.
Now, my attempt at (1)$\implies$(2), inspired by the proof of lemma IV.7.15, is the following: Let $f$ be as above and pick $\dot{f}$ a name for $f$, we wish to see that $f\in M$. For every $\alpha<\theta$, let $e_\alpha\in M$ be such that $M[G]\models \dot{f}(\alpha)=e_\alpha$ and let $A_\alpha=\{p\in\mathbb{P} : p\Vdash \dot{f}(\alpha)=e_\alpha\}$. The problem, of course, is that the $A_\alpha$ need not be dense in $\mathbb{P}$ (indeed, there may be a $q\in \mathbb{P}$ such that $q\Vdash \dot{f}(\alpha)\neq e_\alpha$). But I think this problem could be solved if I take
$A_\alpha=\{p\in\mathbb{P} : p\Vdash \dot{f}(\alpha)=e_\alpha\}\cup\{p\in\mathbb{P} :\exists q\in\mathbb{P}(p\perp q\land q\Vdash \dot{f}(\alpha)=e_\alpha) \}$
and then use (1) to find a element of the intersection that is also in $G$.
However, I have no clue what to do about (2)$\implies$(1) if $\mathbb{P}$ is separative. My initial approach was to take $\{A_\alpha : \alpha < \theta\}$ a family of dense open sets and then assume that there is a $p\in\mathbb{P}$ such that for no element $q\in\bigcap_{\alpha<\theta}A_\alpha$ do we have $q\leq p$. For every $\alpha<\theta$ I would then take $b_\alpha\in A_\alpha$ such that $b_\alpha\leq p$. But I'm not sure how to proceed from there, or where to use the fact that $\mathbb{P}$ is separative.
With respect to the first implication, the family $\langle A_\alpha:\alpha<\theta\rangle$ you defined does not necessarily belong to $M$ because you're using $f$ to define it. Why don't you try instead $A_\alpha:=\{p\in\Bbb P:$ for some $e\in M, p\Vdash \dot f(\check\alpha)=\check e\}$. Prove that $\langle A_\alpha:\alpha<\theta\rangle\in M$.
Now, for the second implication, if $U_\alpha$ is an open dense subset of $\mathbb P$ for each $\alpha<\theta$, choose for all $\alpha<\theta$ an antichain $A_\alpha\subseteq U_\alpha$ and consider the name $\dot f=\{((\check \alpha,\check r),r):\alpha<\theta,r\in A_\alpha\}$.Check that $\Bbb 1\Vdash``\dot f$ is a function from $\check\theta$ into $\check{\Bbb P}"$. Then if you pick an arbitrary $p\in\Bbb P$, prove that there is some $q\leq p$ and $h:\theta\rightarrow\Bbb P$ in $M$ such that $q\Vdash \dot f=\check h$. Then ask yourself why the separability of $\Bbb P$ implies $q\in\bigcap_{\alpha<\theta} U_\alpha$.