Prove that a red-black tree with $n$ internal nodes has height at most $2\lg(n+1)$
The proof seems correct to me.

1) The base case proves the claim when the height of the subtree rooted at $x$ is $0$. The inductive step proves the claim for every subtree rooted at $x$ of positive height $h$ assuming that the claim is true for all subtrees of heights from $0$ to $h-1$. So the inductive step for $h=1$ is able to prove the claim for all subtrees of height $1$ using the fact that the claim is true for the base case, i.e., for all subtrees of height $0$. This is a standard proof by structural induction.

2) Each subtree $T_u$ rooted in a child $u$ of $x$ has a black-height $bh(u)$ of either $bh(x)$ or $bh(x)-1$ but, in any case, the height of $T_u$ is smaller than the height of the subtree $T_x$ rooted in $x$. This means that the induction hypothesis can be applied, showing that the number of internal nodes is at least:


$2^{bh(x)}-1$ if $bh(u)=bh(x)$; or
$2^{bh(x)-1}-1$ if $bh(u)=bh(x)-1$.


In any case the number of internal nodes of $T_u$ is at least the smaller of the above two quantities, i.e., $\min\{ 2^{bh(x)}-1, 2^{bh(x)-1}-1 \} = 2^{bh(x)-1}-1$.
This means that the number of internal nodes in $T_x$ is at least:

$$
2 \cdot ( 2^{bh(x)-1}-1 ) + 1 = 2^{bh(x)}-1.
$$