Which extension of $\sf ZFC$ prove that
$$ {\sf ZFC} \not \vdash \exists x \, ( \operatorname {CH}(x) \land x \neq \emptyset \land x \neq 1)$$


Where $\operatorname {CH}(x) \iff \neg \exists \kappa \, (|x| < \kappa < |P(x)|) $


In English: $\sf ZFC$ doesn't prove the continuum hypothesis of any set other than the empty set and the singleton of the empty set.


I know that "$\sf ZFC + CH$ fails everywhere", can prove that, which is too strong. But I'm asking if this can be proved in a much less strong theory, some theory whose consistency strength is just a little above $\sf ZFC$.



Which extension of ZFC proves that ZFC can only prove CH satisfied by the first two sets?
The statement you propose is equivalent to consistency of GCH failing for all (infinite) cardinals. This problem is well-studied. Of course the theory ZFC+Con(ZFC+"GCH fails everywhere") gives an answer to your problem, but this is completely tautological. It is standard in such situations to calibrate the consistency strength against the large cardinal hierarchy: as is discussed in [this answer](https://math.stackexchange.com/a/3186971/127263), the precise equiconsistency statement is not easy to state in ZFC, but it amounts to existence of stationarily many measurable cardinals of high Mitchell rank. It is between measurable cardinals and strong cardinals.


Let me also mention that your claim in the last paragraph is false: ZFC+"GCH fails everywhere" doesn't prove Con(ZFC+"GCH fails everywhere"), by Godel's incompleteness.


