You are not missing anything. It is true, that every cube contains a rational point, but most squares (or circles) have no rational point. Hence only aleph-zero many cubes.
OTOH, you can divide any cube further to the infinite number of disjunct cubes. Infinitely many times.
Now, if you can organize these divisions in such a way, that those infinities build up to more than aleph-zero cubes, the ZF is down.
I don't say it's possible, it just looks like slightly probable, due to the infinite divisibility of cubes. Which can be misleading, but maybe not.
I'm sorry, didn't you say that the cubes must have no common volume?
If it's worth saying, but not worth its own post, then it goes here.
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