Or can we prove the independence of the CH?
Let's place ourselves in first order logic.
If CH was proved unprovable and ZFC was inconsistent, let P be a paradox in ZFC.
1)We have: P is true
2)And P => CH (since P is false)
3)Therefore CH is true: contradiction..
What's the problem here?
You can certainly prove that ZFC is consistent, but only if you assume some equivalent or stronger theory is consistent. For example, Morse-Kelly set theory proves that ZFC is consistent, but now you're simply transferring the assumption of consistency from ZFC to MK. This is a consequence of Godel's incompleteness theorems. There's no free lunch when it comes to consistency (of sufficiently complex theories).
Now when it comes CH, you are missing a few important details. CH was not proven unprovable, but rather it was proven independent of ZFC. What this means is that it was shown that you can add either CH or $\neg$CH as an additional axiom to ZFC, and assuming that ZFC is consistent, the extended theories ZFC + CH and ZFC + $\neg$CH are also consistent. So CH's independence already relies on the consistency of ZFC, so you cannot use that fact to prove ZFC is consistent.
So when you say
you're basically already at a contradiction since now you're stating that ZFC is both consistent and inconsistent.