Could there be an undecidable statement $S$ in ${\sf ZFC}$ of which one will never be able to prove its undecidability for principal reasons (ie we will never know that $S$ is undecidable)?


If this is a silly question, I apologise; feel free to vote to close. I'll remove it quickly.



Meta-undecidability
One can never prove in ZFC itself that a given statement is independent of ZFC, because the assertion "*$S$ is independent of ZFC*" implies Con(ZFC), since every statement is settled by an inconsistent theory. 


Thus, if a statement $S$ is independent of ZFC, then there is a model of ZFC inside of which $S$ is not independent of ZFC, since this is the situation inside any model of ZFC+$\neg$Con(ZFC). 


For any given computably axiomatizable consistent theory $T$ extending ZFC, there are statements that are independent of $T$ and hence of ZFC, whose ZFC independence cannot be proved in $T$. Namely, the Rosser sentence $\rho$ for $T$. If $T$ is consistent, then $\rho$ is independent of $T$ and hence also of ZFC. But one cannot prove the assertion "*$\rho$ is independent of ZFC*" in the theory $T$, because in a model of T+$\neg$Con(T), there are proofs both of $\rho$ and of $\neg\rho$ in $T$, and depending on which proof has a smaller Gödel code, the model will have either $\rho$ or $\neg\rho$ being true, in a way that even a very weak theory can verify. So $\rho$ will not be independent of ZFC in that model of $T$. 


**Conclusion:** no matter how much you strengthen ZFC, if you do so consistently, there will be ZFC independent statements whose ZFC-independence you cannot establish in your theory.


