Say that a forcing $\mathbb{P}$ is $\sigma$-closed if, for any decreasing sequence $\langle p_n : n \in\omega\rangle$ in P, there is some $q \in \mathbb{P}$ such that $q \leq p_n$ for all $n\in\omega$.
If $\mathbb{P}$ is $\sigma$-closed and $\mathcal{D}$ is a family of dense subsets $|\mathcal{D}|\leq\aleph_1$, then there exists a $\mathbb{P}$-generic over $\mathcal{D}$?
I know, if $\mathbb{P}$ is any notion forcing and $\mathcal{D}$ is a family of dense subsets and $\mathcal{D}:=\langle D_n:n<\omega\rangle$ is countable. We can find a sequence $\langle p_n:n<\omega\rangle$ such that $p_0:=p$, $p_{n+1}\leq p_n$ and $p_n\in D_n$, so set $G:=\{p\in\mathbb{P}:\exists n<\omega(p_n\leq p)\}$ work
I dont use the fact that $\mathbb{P}$ is $\sigma$-closed.
Any advice on how to proceed?
Let $\mathcal D = \{D_\alpha : \alpha <\omega_1\}.$ Recursively construct a decreasing sequence $\langle p_\alpha:\alpha<\omega_1\rangle$ such that $p_\alpha\in D_\alpha.$ The successor stages work just like in the countable case. At a limit stage $\lambda$ take some sequence $\alpha_n$ cofinal in $\lambda$ of type $\omega$ and then use $\sigma$-closedness to get a $q\le p_{\alpha_n}$ for all $n,$ then take $p_\lambda \le q$ such that $p_\lambda \in D_\lambda.$ Like in the countable case, the $p_\alpha$ generate a filter, which is generic by construction.