ccc after strongly proper forcing Let $P, Q \in V$ be such that $P$ is strongly proper and $Q$ is ccc. Does $Q$ continue to be ccc after forcing with $P$?
Since strongly proper forcings do not add new branches to $\omega_1$-trees, they do preserve the ccc-ness of some partial orders: Suslin trees, the poset to specialise an Aronszajn tree. 
 A: Yes.  The proof of Claim 3.8 in Neeman's paper on forcing with side conditions shows that:
(*) if $P$ is strongly proper and $Q$ is proper, $M$ is a countable elementary submodel, $p$ is an $(M,P)$ strong master condition, and $q$ is an $(M,Q)$ master condition, then $p$ forces that $\check{q}$ is an $(M[\dot{G}_P], Q)$-master condition.
(this is an abstraction of a lemma of Sy Friedman)
Now consider also that the following 3 properties of a poset $Q$ are equivalent (this is an old result, I don't recall where, but it's easy to prove):


*

*$Q$ is c.c.c.

*$1_Q$ is a master condition for stationarily many countable models

*$1_Q$ is a master condition for club-many countable models


So back to your question:  assume $P$ is strongly proper and $Q$ is c.c.c.  Then there is a club $C$ of countable models $M$ such that $P$ is strongly proper w.r.t. $M$, and $1_Q$ is an $(M,Q)$ master condition (the latter uses equivalence of 1 with 3).  Let $G_P$ be $(V,P)$-generic.  Using a genericity argument, in $V[G_P]$ the set of $M \in C$ for which $G_P$ includes a strong master condition is stationary; let $S \in V[G_P]$ denote this stationary set.  Also $S':= \{ M[G_P] \ : \ M \in S \}$ is stationary.  Since $S \subseteq C$, then for every $M \in S$, $1_Q$ was an $(M,Q)$-master condition in the ground model.  Then by (*), $V[G_P]$ believes that $1_Q$ is a master condition for every model in the stationary set $S'$.  Hence by the equivalence of 1 and 2 above, $V[G_P]$ believes that $Q$ is c.c.c.
