A tree with $1$ node has height at least $0$, so the claim holds for $n = 1$. Now suppose the claim holds for $n$. Let $T$ be a binary tree with $n + 1$ nodes. Select a leaf node and remove it. By our induction hypothesis, the resulting tree has height at least $\lfloor log_2 n\rfloor$. If $n+1$ is not a power of $2$ then this is equal to $\lfloor log_2 (n+1)\rfloor$, so we are done. Otherwise, choose a leaf of greatest depth to remove. The resulting tree has height at least $\lfloor log_2 n\rfloor$. If the height in fact achieves that, then the only possible tree is the complete binary tree on $n$ vertices. Since every internal node has two children, the only place the removed leaf could have come from is from a leaf vertex. Adding a child to any leaf vertex increases the height of the tree by 1. Since$\lfloor log_2 n\rfloor +1\geq \lfloor log_2 (n+1)\rfloor$, the claim holds.
I don't understand why we should choose a leaf of the greatest depth to remove, and how if the resulting height equals $\lfloor log_2 n\rfloor$ then the only possible tree is a complete binary tree. What if the leaf is not chosen from the greatest depth, or if the height is not equal to $\lfloor log_2 n\rfloor$?