 Solution

False. Take edges $u\to a$ and $c\to u$, and let the outer DFS loop try vertices in order $a,u,c$. The first call finishes $a$ immediately. The next discovers $u$, whose only outgoing edge reaches already black $a$, so its tree contains only $u$. Finally $c$ sees already black $u$.

Vertex $u$ has both an incoming edge from $c$ and an outgoing edge to $a$, yet is the sole vertex of its DFS tree. Nonzero incoming and outgoing degrees do not require either edge to become a tree edge in depth-first search.