Difficulty in understanding a portion in the proof of the $\text{"white path"}$ theorem as with in CLRS text
$\dagger\dagger$ Theorem 22.7 concludes that exactly one of the three conditions listed is satisfied. The inequalities listed in (4) show that the intervals $[d[u],f[u]]$ and $[d[v],f[v]]$ intersect. Therefore, the first condition of the theorem is not the one being satisfied. In the other two conditions one of these two intervals is inside the other. Now, because the inequality $d[u]&lt;d[v]$ is strict, it can't be that $[d[u],f[u]]$ is inside $[d[v],f[v]]$. So, the third condition in the theorem is the only one that can be satisfied.
$\dagger$ This is also using theorem 22.7 or its corollary. The inequality $d[u]&lt;d[v]$ is part of the assumption, that by time $d[u]$ the vertex $v$ is still white. The inequality $f[w]\leq f[u]$ comes from Corollary 22.8, since $w$ is a descendant of $u$. The inequality $d[v]&lt;f[w]$ is due to $(w,v)$ being an edge. Because all neighbors of $w$ must be discovered before we consider it finished. Now, the claim that the proof should end here is jumping to the conclusion of the theorem without justification. It hasn't been justified that DFS will actually use $(w,v)$ to discover $v$.
Example 1: For example, consider the graph with vertices $1,2,3,4$ and (directed) edges $(1,2),(1,4),(2,3),(4,3)$ and we start at $1$ doing DFS. Assume that we already discovered $1$ and $2$.
At this point the vertex $v=3$ is is a white path from $u=w=2$. Just one edge $(2,3)$. DFS will discover $3$ and make it a descendant of $2$. There is no way to change that, since Theorem 22.9 proves it. But imagine that at this point we stop obeying DFS and go to discover $4$ and from $4$ go and discover $3$. Then $3$ wouldn't be a descendant of $2$. It would only be a descendant of $4$ and $1$. Theorem 22.9 proves that DFS wouldn't do that.
Example 2: This is to illustrate the question in the comments. Consider the graph with vertices $1,2,3,4$ and edges $(1,2),(1,4),(2,3),(2,4),(3,4)$. Assume that we have discovered already $1$ and $2$, in that order. The vertex $v=4$ is in a white path from $u=w=2$. However, the edge $(2,4)$ doesn't necessarily need to be used by DFS. If the neighbors of a vertex are considered in the order according to the labels that we gave them in this example, then from $2$ DFS will discover $3$ and from $3$ it will discover $4$. Then $3$ becomes finished, then $2$ becomes finished, and $1$ becomes finished, in that order.
While theorem 22.9 doesn't ensure that either $(2,4)$ or $(3,4)$ will be used, since that depends on the other in which neighbors are explored, note that it does ensure that the edge $(1,4)$ will not be used (under the assumptions of the example that $1$ was discovered and $2$ just got discovered). In fact, Theorem 22.9 implies that $4$ will be a descendant of $2$.