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. 



ccc after strongly proper forcing
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):


1. $Q$ is c.c.c.
2. $1\_Q$ is a master condition for stationarily many countable models
3. $1\_Q$ 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.


