Suppose $\mathbb P$ is a countably closed forcing, and $\mathbb Q$ is a c.c.c. forcing that adds reals. Is $\mathbb P$ still proper in $V^{\mathbb Q}$?
Here is the short version of a negative answer. Let $P$ be the collapse of $\omega_2$ to $\omega_1$ with countable conditions. Fix a tree $(N_\eta:\eta\in 2^{<\omega})$ of quite different models, increasing along each branch. Specifically, make sure that along each branch in $2^\omega$ the union of the models will get a different intersection with $\omega_2$. Now go to any extension $V^Q$ where there is a new real. The new real will give a new branch, yielding a new model; no condition in $P$ can be generic for this model, since a generic condition has to know the model.
(I could not find a counterexample for a week. But since I am in Jerusalem at the moment, I eventually gave up and asked Shelah. It took him only a minute or so to come up with this answer. He also said that the reason he is using this specific $P$ is just to understand generic conditions better; many others would work as well.)
Here is the more explicit version. (Perhaps too explicit? Perhaps too complicated? It is quite possible that I have overlooked a simpler solution.)
Let $P$ be the set of countable partial 11 functions from $\omega_1$ to $\omega_2$. Note that if a condition $q$ is $N$generic it must satisfy the following properties:
 $dom(q) \supseteq \omega_1\cap N $. (Otherwise $q\cup \{(\alpha,\beta)\}$ for some $\alpha\in \omega_1\cap N$, $\beta\in \omega_2\setminus N$ will force that $N[G]$ has the new ordinal $\beta$.)
 $q[\omega_1\cap N]\subseteq \omega_2\cap N$. Obviously.
 $q[\omega_1\cap N] \supseteq \omega_2 \cap N$. (If $\beta\in \omega_2\cap N$ is outside the range of $q[\omega_1\cap N]$, then $q$ would force that $g^{1}(\beta)\in \omega_1\cap N$ is an ordinal outside $N$. Here I write $g$ for the generic function, i.e. the union of all conditions in the filter.)
 Summarizing: $q\restriction(\omega_1\cap N)$ is a bijection between $\omega_1\cap N$ and $\omega_2\cap N$. (If I have not overlooked anything, this is also sufficient for being generic.)
So a generic condition must really know a lot about $N$. We will use this to get a counterexample.
Let $(N_\eta:\eta\in 2^{<\omega})$ be a family of countable elementary submodels with the following properties:
 For alle $\eta\subseteq \nu$ we have $N_\eta \subseteq N_\nu$.
 For all $\eta,\nu\in 2^\omega$ with $\eta\not=\nu$ we have $N_\eta \cap \omega_2 \not=N_\nu\cap \omega_2$. ($N_\eta$ is defined naturally as $\bigcup_n N_{\eta\restriction n}$.)
 Moreover: For every $\eta\in 2^{<\omega}$ there is an ordinal $\alpha_\eta\in \omega_2$ witnessing that the two branches that split at $\eta$ will yield models whose intersections with $\omega_2$ are different: $\alpha_\eta\in N_{\eta^\frown 0}$, but not in $N_{\eta^\frown1}$, and also $\alpha_\eta\notin N_\nu$ for any $\nu$ extending $\eta^\frown1$.
 There is a $\delta$ such that $\omega_1\cap N_\eta=\delta$ for all $\eta\in 2^\omega$. (We will in fact have $\omega_1\cap N_\eta=\delta$ also for all $\eta\in 2^{<\omega}$.)
Now let $Q$ be a (ccc, or just proper) forcing adding a new real $\nu\in 2^\omega$. In the extension $V^Q$ the family $(N_{\nu\restriction n}[G]:n\in \omega)$ is an increasing sequence of elementary submodels; its union $N_\nu[G]$ is again an elementary submodel, and $N_\nu[G]\cap \omega_1=\delta$.
Assume that $q$ is generic for this model. Then $S:=q[\delta]=\omega_2 \cap N_\nu[G]$ is a set in the ground model. But now $\nu$ can be computed from $S$: $\eta(0)=1 $ iff $\alpha_{\langle\rangle}\in S$, $\eta(1)=1 $ iff $\alpha_{\langle\eta(0)\rangle}\in S$, etc. Contradiction.
It remains to find a tree as described above. (I took some of the following arguments from chapter XV of "Proper and Improper Forcing". I suspect there might be a much easier construction.)
 Start with a Namba tree $(N_s:s\in \omega_2^{<\omega})$ of countable models which are increasing along each branch and satisfy $s\in N_s$ for all $s\in \omega_2^{<\omega}$.
 Thin out the tree to make all $N_s\cap\omega_1$ equal to some $\delta$ for all $s\in \omega_2^\omega$, still keeping $\aleph_2$ many successors of each node. (Proper and improper, XV 2.12, based on XI 3.5, going back to a partition theorem in Rudin+Shelah, APAL 1987, RuSh:117. Use determinacy of a low level Borel game.)
 Replace each model $N_s$ ($s\in \omega_2^{<\omega}$) by the Skolem closure of $N_s\cup\delta$.
 Now all $\omega_1\cap N_s=\delta$ for all $s\in \omega_2^{\le \omega}$.
Now find a binary tree $(s_\eta:\eta\in 2^{<\omega})$ with corresponding models $M_\eta:=N_{s_\eta}$ as follows.
 Start with $s_{\langle\rangle}:=\langle\rangle$, $M_{\langle\rangle}=N_{\langle\rangle}$.
 Given $M_\eta=N_{s_\eta}$, first pick an ordinal $\alpha_\eta$ which appears in one of the successors of $M_\eta$ in the $N$tree (i.e., in some $N_{s_\eta^\frown i}$, but not in coboundedly many of them).
 Let $M_{\eta^\frown 0}$ be a successor of $M_{\eta}$ containing $\alpha_\eta$.
 Let $M_{\eta^\frown 1}$ be a successor of $M_{\eta}$ not containing $\alpha_\eta$, but containing some larger ordinal $\beta_\eta<\omega_2$.
 Note that no model $M\supseteq M_{\eta^\frown1}$ with $M\cap \omega_1=\delta$ can contain $\alpha_\eta$ as an element, since already $M_{\eta^\frown1}$ sees a bijection between $\omega_1$ and $\beta_\eta$; if a new element appeared in the range of this bijection, a new element would have to appear also in the domain.
This finishes the inductive step, and thus completes the construction of the tree of models.

1$\begingroup$ "It took him only a minute or so to come up with this answer."  That long, huh? :) $\endgroup$ – Asaf Karagila♦ Mar 6 '15 at 22:41

3$\begingroup$ Literally a minute or two. (And I really mean "literally" in the literal sense.) I think that some of this time he spent deliberating whether a single forcing $P$ would be enough for him (or me), or whether he should go for a claim "all nontrivial $\sigma$closed $P$..." $\endgroup$ – Goldstern Mar 7 '15 at 13:49

1$\begingroup$ @Goldstern I believe you mean Rubin+Shelah, 'Combinatorial properties on trees...'? Also, I'm not sure I understand the last sentence of your comment. Did either of you consider this claim (but find a counterexample/think it plausible but not dwell too much on it/am I misunderstanding completely)? $\endgroup$ – tci Mar 9 '15 at 12:26

1$\begingroup$ @TanmayInamdar Matti Rubin of course, not Rudin. Thank you for the correction.  I think it is plausible that any nontrivial $\sigma$closed forcing can serve as a counterexample (assuming nonCH or at least that there is no dense set of size $\aleph_1$ below any condition  we need this assumption by Victoria Gitman's answer), but I have not thought about it much. $\endgroup$ – Goldstern Mar 9 '15 at 13:48
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 $\omega_1$.

$\begingroup$ To clarify the "removal" of CH: If $P$ is $\sigma$closed and nontrivial (say: corresponds to an atomless Boolean algebra), then $P$ has size at least continuum. So you still need CH. (You just don't need to mention it as it follows from $P=\aleph_1$.) $\endgroup$ – Goldstern Mar 1 '15 at 11:14

$\begingroup$ Ah, yes, of course, ${\rm CH}$ still has to hold, so the assumption is implicit. $\endgroup$ – Victoria Gitman Mar 1 '15 at 14:45