Can someone explain why, under certain conditions, a product forcing $\mathbb P \times \mathbb Q$ satisfies ccc. property.
Namely, if $\mathbb P$ and $\mathbb Q$ are two forcings and if the following holds:
$\mathbb P$ is ccc. and $1_{\mathbb P}\Vdash_{\mathbb P}$"$\overset {\vee} {\mathbb Q}$ is ccc."
then $\mathbb P \times \mathbb Q$ is ccc.
Here $\overset {\vee} {\mathbb Q}$ is just the canonical name.
Suppose, toward a contradiction, that you had an $\omega_1$-sequence of pairwise incompatible elements $(p_\alpha,q_\alpha)$ in $\mathbb P\times\mathbb Q$. Notice that $$ A=\{(p_\alpha,\check q_\alpha):\alpha<\omega_1\} $$ is a $\mathbb P$-name for a subset of $\check{\mathbb Q}$.
I claim that every condition in $\mathbb P$ forces "$A$ is an antichain in $\check{\mathbb Q}$"; more precisely, "If $q_\alpha$ and $q_\beta$ are both in $A$ and $\alpha\neq\beta$, then they are incompatible." (Strictly speaking the subscripts on $q$ should be $\check\alpha$ and $\check\beta$, but I'm too lazy for that.) To see this, suppose not. So some condition $z\in\mathbb P$ must force "$\check q$ compatible with $\check q'$" for some distinct elements $q,q'\in\mathbb Q$ (more precisely, distinct subscripts in the enumeration of the $q_\alpha$'s). Using the definition of how membership statements are forced, we get an extension of $z$ that also extends $p_\alpha$ for some $\alpha$ such that $q=q_\alpha$, and then we get a further extension that also extends $p_\beta$ for some $\beta$ such that $q'=q_\beta$. So $p_\alpha$ and $p_\beta$ are compatible. Since $(p_\alpha,q_\alpha)$ and $(p_\beta,q_\beta)$ are incompatible in $\mathbb P\times\mathbb Q$, it follows that $q_\alpha$ and $q_\beta$ are incompatible in $\mathbb Q$. Thus, we have an extension of $z$ forcing "$q$ and $q'$ are incompatible," contrary to our choice of $z$. This contradiction completes the proof of the claim that $A$ is an antichain in the forcing extension.
Since $\mathbb P$ forces that $\check{\mathbb Q}$ is ccc, it must also force that $A$ is countable. Furthermore, since it's obviously forced that $A\subseteq\{q_\alpha:\alpha<\check\omega_1\}$, and since $\omega_1$ is absolute for $\mathbb P$-forcing thanks to the ccc for $\mathbb P$, we conclude that $\mathbb P$ forces $(\exists\gamma<\check\omega_1)(\forall\alpha<\omega_1)\,(q_\alpha\in A\implies\alpha<\gamma)$. Choose a maximal antichain $B\subseteq\mathbb P$ of conditions $p$ deciding particular values $\check\gamma_p$ for the smallest such $\gamma$. Because $\mathbb P$ is ccc, $B$ is countable, so we can fix a countable ordinal $\delta$ greater than all of the $\gamma_p$ for $p\in B$.
By maximality of $B$, $p_\delta$ is compatible with some $p\in B$. But $p_\delta$ and $p$ force contradictory statements: $p_\delta$ forces $\check q_\delta\in A$, by definition of $A$, whereas $p$ forces that whenever $\check q_\alpha\in A$ then $\alpha<\gamma_p<\delta$. This contradiction completes the proof that $\mathbb P\times\mathbb Q$ cannot have an $\omega_1$-sequence of pairwise incompatible elements.
EDIT: I think I wasn't careful enough about the possibility that $q_\alpha=q_\beta$ even when $\alpha\neq\beta$, so let me point out the following. Consider the (possibly many) $\alpha$'s for which $q_\alpha$ has some particular value $q$. The corresponding $p_\alpha$'s have to be incompatible in $\mathbb P$ because the corresponding pairs $(p_\alpha,q_\alpha)$ are incompatible in $\mathbb P\times\mathbb Q$. By the ccc for $\mathbb P$, there can only be countably many such $\alpha$'s. In other words, the issue that I wasn't careful enough about causes only countable-sized problems. With this observation, what I wrote yesterday should work with no (further) problems.