$v$ is not a descendant of $u \implies u$ was already non-white when visit(w) was called. Therefore, $s(v) < s(w)$.
I couldn't get it why $s(v) < s(w)$ holds. Since $w$ is the predecessor of $v$ in the white path, it seems that $s(v) > s(w)$ is the case. Please let me know if I'm wrong. Besides the $s(v) < s(w)$, I also couldn't understand how “ $v$ is not a descendant of $u \implies u$ was already non-white when visit(w) was called ” holds.
Thanks for reporting this issue! I couldn't understand what I had written, so I rewrote the proof. Let me know if you still have issues with the proof.
https://sharmaeklavya2.github.io/theoremdep/nodes/graph-theory/dfs/white-path-theorem.html
I couldn't get it why $s(v) < s(w)$ holds. Since $w$ is the predecessor of $v$ in the white path, it seems that $s(v) > s(w)$ is the case. Please let me know if I'm wrong. Besides the $s(v) < s(w)$, I also couldn't understand how “ $v$ is not a descendant of $u \implies u$ was already non-white when
visit(w)
was called ” holds.