I just meant that if $|A|=|\cup x|$, meaning they are equinumerous, then $A$ is equinumerous with $\cup x$, and so we can replace $x$ with $x'$ where $A=\cup x'$. So that part of your change didn't actually change anything. Meanwhile, the rest of my answer here does use that $\omega_1$ is well-orderable, so that all the cardinals below it are also well-orderable, and this is how I concluded that $x$ must be countable and that the elements of $x$ must be countable as well, and indeed finite for the hypersingular case.