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I have a very partial answer to the question. I suspect that the argument given in the proof of Theorem 3.5 in my paper Proper and piecewise proper families of reals generalizes to show that if $\mathbb P$ is countably closed and has size $\omega_1$, then it remains proper after forcing with the ccc poset $\mathbb Q$. Here is a sketch.

Let's argue that $\mathbb P$ is still proper in a forcing extension $V[g]$ by $\mathbb Q$. Decompose $\mathbb P$ as an increasing chain $\mathbb P_\xi$ of countable subposets. For a large enough $\lambda$, we need to show that there is a club of countable $N\prec H_\lambda^{V[g]}$ for which $\mathbb P$ has generic conditions. We can assume that $H_\lambda^{V[g]}=H_\lambda[g]$. Consider the club of models of the form $N=M[g]$, where $M\subseteq V$, $M\prec H_\lambda$ (althought it may not be an element of $V$ and the sequence $\{\mathbb P_\xi\mid\xi<\omega_1\}$ is an element of $M[g]$. It follows that $M[g]\cap \mathbb P=\mathbb P_\xi$ for some $\xi$ is an element of $V$. Consider the countably many dense subsets of $\mathbb P$ that are elements of $M$ (not $M[g]$). Since each of them is in $V$ and $\mathbb Q$ is ccc, we can cover them by a countable set of dense sets in $V$ and therefore we can find an $M$-generic condition $q$. But in fact we will argue that $q$ is $M[g]$ generic.

Let $G$ be some $\mathbb P$ generic for $V[g]$ containing $q$. It suffices to show that $M[g][G]$ and $M[g]$ have the same ordinals. Note that $M[g][G]=M[G][g]$ and $M[G]$ and $M$ have the same ordinals. So it remains to show that $M[G]$ and $M[G][g]$ have the same ordinals, meaning that $g$ is $M[G]$-generic. Note that $M[G]\prec H_\lambda[G]$. Let $A$ be an antichain of $\mathbb Q$ in $M[G]$, then $A\in H_\lambda[G]$ and $A$ is countable (because $\mathbb Q$ remains ccc after countably closed forcing). Thus, $A$ is contained in $M[G]$, which means that $g$, which is $V[G]$-generic, meets it and thus $g$ is $M[G]$-generic.

I have a very partial answer to the question. I suspect that the argument given in the proof of Theorem 3.5 in my paper Proper and piecewise proper families of reals generalizes to show that if $\mathbb P$ is countably closed and has size $\omega_1$, then it remains proper after forcing with the ccc poset $\mathbb Q$. Here is a sketch.

Let's argue that $\mathbb P$ is still proper in a forcing extension $V[g]$ by $\mathbb Q$. Decompose $\mathbb P$ as an increasing chain $\mathbb P_\xi$ of countable subposets. For a large enough $\lambda$, we need to show that there is a club of countable $N\prec H_\lambda^{V[g]}$ for which $\mathbb P$ has generic conditions. We can assume that $H_\lambda^{V[g]}=H_\lambda[g]$. Consider the club of models of the form $N=M[g]$, where $M\subseteq V$, $M\prec H_\lambda$ (althought it may not be an element of $V$ and the sequence $\{\mathbb P_\xi\mid\xi<\omega_1\}$ is an element of $M[g]$. It follows that $M[g]\cap \mathbb P=\mathbb P_\xi$ for some $\xi$ is an element of $V$. Consider the countably many dense subsets of $\mathbb P$ that are elements of $M$ (not $M[g]$). Since each of them is in $V$ and $\mathbb Q$ is ccc, we can cover them by a countable set of dense sets in $V$ and therefore we can find an $M$-generic condition $q$. But in fact we will argue that $q$ is $M[g]$ generic.

Let $G$ be some $\mathbb P$ generic for $V[g]$ containing $q$. It suffices to show that $M[g][G]$ and $M[g]$ have the same ordinals. Note that $M[g][G]=M[G][g]$ and $M[G]$ and $M$ have the same ordinals. So it remains to show that $M[G]$ and $M[G][g]$ have the same ordinals, meaning that $g$ is $M[G]$-generic. Note that $M[G]\prec H_\lambda[G]$. Let $A$ be an antichain of $\mathbb Q$ in $M[G]$, then $A\in H_\lambda[G]$ and $A$ is countable (because $\mathbb Q$ remains ccc after countably closed forcing). Thus, $A$ is contained in $M[G]$, which means that $g$, which is $V[G]$-generic, meets it and thus $g$ is $M[G]$-generic.

Edit: I removed the ${\rm CH}$ assumption, which I realized is replaced by the assumption that $\mathbb P$ has size $\omega_1$.

I have a very partial answer to the question. I suspect that the argument given in the proof of Theorem 3.5 in my paper Proper and piecewise proper families of reals generalizes to show that under CH, if $\mathbb P$ is countably closed and has size $\omega_1$, then it remains proper after forcing with the ccc poset $\mathbb Q$. Here is a sketch.

Let's argue that $\mathbb P$ is still proper in a forcing extension $V[g]$ by $\mathbb Q$. Decompose $\mathbb P$ as an increasing chain $\mathbb P_\xi$ of countable subposets. For a large enough $\lambda$, we need to show that there is a club of countable $N\prec H_\lambda^{V[g]}$ for which $\mathbb P$ has generic conditions. We can assume that $H_\lambda^{V[g]}=H_\lambda[g]$. Consider the club of models of the form $N=M[g]$, where $M\subseteq V$, $M\prec H_\lambda$ (althought it may not be an element of $V$ and the sequence $\{\mathbb P_\xi\mid\xi<\omega_1\}$ is an element of $M[g]$. It follows that $M[g]\cap \mathbb P=\mathbb P_\xi$ for some $\xi$ is an element of $V$. Consider the countably many dense subsets of $\mathbb P$ that are elements of $M$ (not $M[g]$). Since each of them is in $V$ and $\mathbb Q$ is ccc, we can cover them by a countable set of dense sets in $V$ and therefore we can find an $M$-generic condition $q$. But in fact we will argue that $q$ is $M[g]$ generic.

Let $G$ be some $\mathbb P$ generic for $V[g]$ containing $q$. It suffices to show that $M[g][G]$ and $M[g]$ have the same ordinals. Note that $M[g][G]=M[G][g]$ and $M[G]$ and $M$ have the same ordinals. So it remains to show that $M[G]$ and $M[G][g]$ have the same ordinals, meaning that $g$ is $M[G]$-generic. Note that $M[G]\prec H_\lambda[G]$. Let $A$ be an antichain of $\mathbb Q$ in $M[G]$, then $A\in H_\lambda[G]$ and $A$ is countable (because $\mathbb Q$ remains ccc after countably closed forcing). Thus, $A$ is contained in $M[G]$, which means that $g$, which is $V[G]$-generic, meets it and thus $g$ is $M[G]$-generic.

I have a very partial answer to the question. I suspect that the argument given in the proof of Theorem 3.5 in my paper Proper and piecewise proper families of reals generalizes to show that if $\mathbb P$ is countably closed and has size $\omega_1$, then it remains proper after forcing with the ccc poset $\mathbb Q$. Here is a sketch.

Let's argue that $\mathbb P$ is still proper in a forcing extension $V[g]$ by $\mathbb Q$. Decompose $\mathbb P$ as an increasing chain $\mathbb P_\xi$ of countable subposets. For a large enough $\lambda$, we need to show that there is a club of countable $N\prec H_\lambda^{V[g]}$ for which $\mathbb P$ has generic conditions. We can assume that $H_\lambda^{V[g]}=H_\lambda[g]$. Consider the club of models of the form $N=M[g]$, where $M\subseteq V$, $M\prec H_\lambda$ (althought it may not be an element of $V$ and the sequence $\{\mathbb P_\xi\mid\xi<\omega_1\}$ is an element of $M[g]$. It follows that $M[g]\cap \mathbb P=\mathbb P_\xi$ for some $\xi$ is an element of $V$. Consider the countably many dense subsets of $\mathbb P$ that are elements of $M$ (not $M[g]$). Since each of them is in $V$ and $\mathbb Q$ is ccc, we can cover them by a countable set of dense sets in $V$ and therefore we can find an $M$-generic condition $q$. But in fact we will argue that $q$ is $M[g]$ generic.

Let $G$ be some $\mathbb P$ generic for $V[g]$ containing $q$. It suffices to show that $M[g][G]$ and $M[g]$ have the same ordinals. Note that $M[g][G]=M[G][g]$ and $M[G]$ and $M$ have the same ordinals. So it remains to show that $M[G]$ and $M[G][g]$ have the same ordinals, meaning that $g$ is $M[G]$-generic. Note that $M[G]\prec H_\lambda[G]$. Let $A$ be an antichain of $\mathbb Q$ in $M[G]$, then $A\in H_\lambda[G]$ and $A$ is countable (because $\mathbb Q$ remains ccc after countably closed forcing). Thus, $A$ is contained in $M[G]$, which means that $g$, which is $V[G]$-generic, meets it and thus $g$ is $M[G]$-generic.

I have a very partial answer to the question. I suspect that the argument given in the proof of Theorem 3.5 in my paper Proper and piecewise proper families of reals generalizes to show that under CH, if $\mathbb P$ is countably closed and has size $\omega_1$, then it remains proper after forcing with the ccc poset $\mathbb Q$. Here is a sketch.

Let's argue that $\mathbb P$ is still proper in a forcing extension $V[g]$ by $\mathbb Q$. Decompose $\mathbb P$ as an increasing chain $\mathbb P_\xi$ of countable subposets. For a large enough $\lambda$, we need to show that there is a club of countable $N\prec H_\lambda^{V[g]}$ for which $\mathbb P$ has generic conditions. We can assume that $H_\lambda^{V[g]}=H_\lambda[g]$. Consider the club of models of the form $N=M[g]$, where $M\subseteq V$, $M\prec H_\lambda$ (althought it may not be an element of $V$ and the sequence $\{\mathbb P_\xi\mid\xi<\omega_1\}$ is an element of $M[g]$. It follows that $M[g]\cap \mathbb P=\mathbb P_\xi$ for some $\xi$ is an element of $V$. Consider the countably many dense subsets of $\mathbb P$ that are elements of $M$ (not $M[g]$). Since each of them is in $V$ and $\mathbb Q$ is ccc, we can cover them by a countable set of dense sets in $V$ and therefore we can find an $M$-generic condition $q$. But in fact we will argue that $q$ is $M[g]$ generic.

Let $G$ be some $\mathbb P$ generic for $V[g]$ containing $q$. It suffices to show that $M[g][G]$ and $M[g]$ have the same ordinals. Note that $M[g][G]=M[G][g]$ and $M[G]$ and $M$ have the same ordinals. So it remains to show that $M[G]$ and $M[G][g]$ have the same ordinals, meaning that $g$ is $M[G]$-generic. Note that $M[G]\prec H_\lambda[G]$. Let $A$ be an antichain of $\mathbb P$ in $M[G]$, then $A\in H_\lambda[G]$ and $A$ is countable (because $\mathbb Q$ remains ccc after countably closed forcing). Thus, $A$ is contained in $M[G]$, which means that $g$, which is $V[G]$-generic, meets it and thus $g$ is $M[G]$-generic.

I have a very partial answer to the question. I suspect that the argument given in the proof of Theorem 3.5 in my paper Proper and piecewise proper families of reals generalizes to show that under CH, if $\mathbb P$ is countably closed and has size $\omega_1$, then it remains proper after forcing with the ccc poset $\mathbb Q$. Here is a sketch.

Let's argue that $\mathbb P$ is still proper in a forcing extension $V[g]$ by $\mathbb Q$. Decompose $\mathbb P$ as an increasing chain $\mathbb P_\xi$ of countable subposets. For a large enough $\lambda$, we need to show that there is a club of countable $N\prec H_\lambda^{V[g]}$ for which $\mathbb P$ has generic conditions. We can assume that $H_\lambda^{V[g]}=H_\lambda[g]$. Consider the club of models of the form $N=M[g]$, where $M\subseteq V$, $M\prec H_\lambda$ (althought it may not be an element of $V$ and the sequence $\{\mathbb P_\xi\mid\xi<\omega_1\}$ is an element of $M[g]$. It follows that $M[g]\cap \mathbb P=\mathbb P_\xi$ for some $\xi$ is an element of $V$. Consider the countably many dense subsets of $\mathbb P$ that are elements of $M$ (not $M[g]$). Since each of them is in $V$ and $\mathbb Q$ is ccc, we can cover them by a countable set of dense sets in $V$ and therefore we can find an $M$-generic condition $q$. But in fact we will argue that $q$ is $M[g]$ generic.

Let $G$ be some $\mathbb P$ generic for $V[g]$ containing $q$. It suffices to show that $M[g][G]$ and $M[g]$ have the same ordinals. Note that $M[g][G]=M[G][g]$ and $M[G]$ and $M$ have the same ordinals. So it remains to show that $M[G]$ and $M[G][g]$ have the same ordinals, meaning that $g$ is $M[G]$-generic. Note that $M[G]\prec H_\lambda[G]$. Let $A$ be an antichain of $\mathbb Q$ in $M[G]$, then $A\in H_\lambda[G]$ and $A$ is countable (because $\mathbb Q$ remains ccc after countably closed forcing). Thus, $A$ is contained in $M[G]$, which means that $g$, which is $V[G]$-generic, meets it and thus $g$ is $M[G]$-generic.