Your argument is correct, but there are plenty of nonempty sets with zero measure e.g. Cantor sets.

@fedja I have laid my hand on Kahane's book. In particular, it contains a much better proof of the case $A > 1$. Thanks for the reference!

Unfortunately, some of the reference books on this topic are behind paywalls, so I couldn't study them...

More generally, this is called "Dvoretzky problem". It looks like the 1D case (covering interval with smaller intervals) is completely solved, and if you believe the 2D case is the same then the answer to this question should be "yes". However, I couldn't find a reference for the 2D case.

Thank you for the references!

I have posted another question here mathoverflow.net/questions/442538/…. Feel free to take a look :)