A proof about $\sigma$-algebras via transfinite induction This is a proofreading question.

I was trying to help out on this question and in the course of that I encountered the following assertion:

Let $(X, \mathcal A)$ and $(Y, \mathcal B)$ be $\sigma$-algebras, and let $(X\times Y, \mathcal A \otimes \mathcal B)$ be their product $\sigma$-algebra. Denote by $\pi_X$ the projection $X \times Y \to X$.
Then for any $S \in \mathcal A \otimes \mathcal B$, the set $\forall_\pi(S) := \{x \in X \mid \forall y \in Y: (x,y) \in S\}$ is measurable.

Now we know that $\mathcal A \otimes \mathcal B = \sigma(\mathcal A \times \mathcal B)$, and the latter may be constructed from $\mathcal A \times \mathcal B$ by transfinite induction on $\aleph_1$, as set out in $\S 3$ of this paper.
Thus we can verify that the assertion above holds by transfinite induction, starting at the generators $\mathcal A \times \mathcal B$ (for which it is trivial) and establishing "closure of the assertion" under countable union and complement (and countable intersection, if you like to "speed things up"). Verification at limit ordinals is trivial.

Now this outline seems to prove the assertion. The problem I have is that the source for ref'd question, Set-valued analysis by Aubin-Frankowska, states with regard to the there-requested theorem:

It is quite tempting to think that the projection [...] of a Borel set should be still a Borel set. [...] The discovery of Borel sets whose continuous image is not Borel started the intensive study of analytic sets, i.e., continuous images of Borel sets. [...]

and references Convex analysis and measurable multifunctions by Castaing, where a very technical and non-illuminating proof is given that I don't understand at the moment. But in mathematics chat, I and @Vrouvrou have reduced this "Measurable Projection Theorem" to a special case of above asssertion.

I find it quite hard to believe that this approach would discard (part of) the motivation for a whole area of research (viz on analytic sets) and surpass a hard, technical proof. So, where is my mistake?
 A: As I suspected, there was an error. It is not surprising that none of the others were able to find it, because I hid it in the small phrase:

[...] establishing "closure of the assertion" under countable union [...]

This is not true. Suppose we are at step $\alpha$, and we have a countable collection $(S_{\alpha,n})_{n \in \Bbb N}$ of sets in $\sigma_\alpha(\mathcal A \times \mathcal B)$. Then we can manipulate a bit as follows:
$$\forall_\pi\left(\bigcup S_{\alpha,n}\right) = \pi_X\left[\left(\bigcup S_{\alpha,n}\right)^c\right]^c = \pi_X \left[\bigcap S_{\alpha,n}^c\right]^c$$
but in general it may occur that e.g.:
$$\pi_X\left[\bigcap S_{\alpha,n}^c\right] \subsetneq \bigcap \pi_X[S_{\alpha,n}^c]$$
Thus this train of thought does not follow through.

Rather, while this was invented to prove that $\pi_X(S)\in \mathcal A \implies \pi_X(S^c) \in \mathcal A$, it becomes reasonable to believe that extra conditions on the measure spaces may be necessary. This is in fact in line with the observations made in the question statement.
