Prove that any subtree in a red-black tree has at least $2^{bh(x)} -1$ internal nodes
It turns out trivial. I think I found the answer myself. First of all, $bh(x)$ gives us the black-height rooted from the node x. A subtree rooted from node x, in the worst case (in terms of the number of nodes it has) should not include any red nodes. Any node should be there in order to satisfy the $bh(x)$ number and nothing more. In that case we would boil down to a perfect subtree of the height $bh(x)$ and we now that a perfect binary tree has $2^{bh(x)}-1$ nodes. This is the least number of nodes it can have.
The subtlety in the question is that since we are considering the child of a node x that has $bh(x)$, it has at least $2^{bh(x)-1}-1$ nodes due to the possibility of losing 1 black-height by walking from x to its child.
PS: Even though what i wrote above seems sensible to me and maybe gives insight to the lemma even without the need of induction, after a little bit of careful reading, that claim is nothing but the just the induction hypothesis that holds for the leafs and propagates to the top of the three.