This is a HW question. The conditions seem to be talking about a cycle graph (they are 2 connected and have vertices of degree 2). For them the condition checks out. But I don't know for sure it refers to them.
My next thought was to try a proof by contradiction. So we assume u is not on one the paths.It has to be part of V(G) since the graph is 2 connected. I feel like the degree is significant, but I don't know where to go. Do we say something like u has to be connected to some vertices on some of the paths, making it part? How do I prove that claim? Am I totally wrong?
This seems to be a solution that disproves the thing. It is 2 connected and the deg(u) is 2. It's also not on any ear decomp path because if you go to u, you end up in another one. Is this allowed, since we restrict the Paths themselves?