Unprovable unprovability In general, mathematical conjectures are resolved by proof, disproof, or proof that they are neither provable nor disprovable.
Is it possible that some open conjectures cannot be settled in any of these ways? That is, are there statements that are independent of, say, $\mathrm{ZFC}$, but not provably independent?
 A: There is always in increase in strength, relative to a fixed theory, to prove that something is unprovable in that theory. It follows from the incompleteness theorems that for a sufficiently strong effective theory $T$ (say ZFC), if $T$ proves "$\phi$ is not provable in $T$" for one sentence $\phi$, then $T$ proves $\text{Con}(T)$ and $T$ is inconsistent. 
So whenever we want to prove "$\phi$ is not provable in $T$", we have to do so by means not formalizable in $T$. We can add an assumption such as $\text{Con}(T)$, or we can extend $T$ in other ways, for example if $T$ is Peano arithmetic we might prove the independence result in ZFC. 
Thus the way that we know a sentence to be independent of ZFC is by proving that independence in some stronger theory. So, while there may be statements independent of ZFC which we have not yet proved are independent of ZFC, we will not know they are independent, so we will not be able to give them as concrete examples for the question being asked. 
It is possible to give plenty of examples of particular theories $S$ extending ZFC, and sentences $\phi$ independent of ZFC, so that $S$ cannot prove that $\phi$ is independent of ZFC. The easiest way is to take $\phi$ to be $\text{Con}(S)$ for some theory $S$ whose consistency is not provable in ZFC, e.g. letting $S$ be ZFC plus the existence of a measurable cardinal. I would be interested to see whether other answerers can give more natural examples. 
