I'm attempting to understand the answer to the question, Is there a statement whose undecidability is undecidable (as in independent, not a decision problem)? The answer appears to be "Yes". However, in both top-voted answers, the answer depends on this:
Specifically, it's impossible for ZFC to prove that it can't prove something, because such a statement is tantamount to the consistency of ZFC; if ZFC were inconsistent then it could prove everything, so proving that there's something that can't be proved is equivalent to proving the consistency of ZFC, and of course this is forbidden by the second incompleteness theorem.
At first I didn't understand this, but now I think I do: If ZFC is inconsistent, then ZFC could prove anything. Thus, proving that it can't prove something proves that ZFC is consistent, since if it were inconsistent then it could prove everything. And if ZFC can prove that it is consistent, then ZFC is inconsistent, as per Godel's second incompleteness theorem.
However, there are statements known to be undecidable in ZFC. If a statement $S$ is known to be undecidable, is this not a proof that $(ZFC \not\vdash S) \land (ZFC \not\vdash \neg S)$ - meaning ZFC shows it can't prove something (two things, to be exact)?
Or is it that the proof of undecidability is not within ZFC and so ZFC is safe?