Prove that exists Gray Codes of length $\lceil \log_2 k \rceil$ for any positive integer $k$. The Gray codes for even $k$ values are closed (form a unique cycle), the ones for odd $k$ values are open.
The inductive hypothesis is that exists Gray codes of length $\lceil \log_2 k \rceil$ for all values $k<n$. If $k$ is even, then the code is closed, otherwise it's open.
First, the book says the base case is trivial. However, for $k=1$ one would have a Gray code of length $0$ which is a bit strange.
Despite that, I understood the proof of the first particular case where $n$ is even. It's just take a Gray code of length $\frac{n}{2}$ and make 2 copies, one with leading zeros and the other with leading ones. If we write them down, it's possible to see that arranging them we can form a cycle. Moreover, we added 1 bit to the code and got a $n$ length new code, so the number of bits of the new code is $\lceil \log_2 \frac{n}{2} \rceil + 1=\lceil \log_2 n \rceil$.
When n is odd I do not understood the proof given. It is as follows:
Let $n=2k+1$. Construct two Gray codes of size $k$, and connect them in the same way as before. If $2k$ is not a power of $2$, then there are some strings with length $\lceil \log_2 2k \rceil$, which have not been used as names. One of these strings is connected to one of the strings that has been used. We can now break the cycle of length $2k$ by adding this new string, resulting in an open path of length $2k+1$. The number of bits satisfies the condition. If $2k$ is a power of 2, there are no unused strings left, and we need to add one more bit to the code. The total number of bits is thus $\lceil \log_2 2k \rceil + 1=\lceil \log_2 (2k+1) \rceil$.
I don't comprehend what it means by unused strings when $2k$ is not a power of 2 or no unused strings for $2k$ being a power of 2.