I know that the continuum hypothesis is not decidable, i.e. we can not prove it, nor disprove it.
The question is, is it theoretically possible to find an explicit set $E\subset \mathbb R$ such that we can not prove neither $\vert \mathbb N\vert =\vert E\vert$, nor $\vert \mathbb R\vert =\vert E\vert$?
Have you ever heard of such a set before?
This has nothing to do with the continuum hypothesis, but it is quite possible. Let $S(n)$ be a predicate of natural numbers $n$ such that neither "there is some $n$ such that $S(n)$" nor "there is no $n$ such that $S(n)$" is provable. Let $E = \mathbb Z \cup \{(n, n+1):\; S(n)\}$. Thus if there is some $n$ such that $S(n)$, $E$ has cardinality of the continuum, while if there is no such $n$, $E$ is countable.