We just saw this theorem in class. When proving this implication, however:
$$G \text{ is a $2$-connex graph } \Rightarrow \forall x,y \in V_G \text{ exist two internally disjoint paths} $$
I thought the proof to be overly complicated, so I was thinking if this proof was correct:
Let $G$ be a 2-connex graph and suppose there is a pair of vertexes such that there are no internally disjoint paths between them.
But this means that every path between them shares at least one vertex, which we'll call $v$. So $v$ is a cut vertex. But if $G$ has a cut vertex then $k_v(G) = 1$, which contradicts the fact that $G$ is a 2-connex graph
I asked my teacher if this proof was correct but she thinks there must be something missing else the proof she gave us wouldn't be so cumbersome, but she says she's not sure what could be missing.
That's why I'm asking whether I'm missing something or if this is fine.
Thanks in advance