Let $G$ be a 2-connected simple graph. Let $s$ and $t$ be vertices in $G$. Prove that the vertices of $G$ can be linearly ordered so that each vertex apart from $s$ and $t$ has a neighbor that is earlier in the order and a neighbor that is later in the order. (Hint: use ear decompositions)
My thought process: use induction on the number of ears. If the number of ears is one, then we just have a cycle and for any $s$ and $t$ on the cycle I make $s=1$ (i.e. has index 1 in the order) and every vertex to the right of $s$ gets indices 2,3,...etc one by one moving to the right of $s$ until I reach $t$ which I will index as $t=n$ ($n$ being the number of vertices in the cycle). Then I keep moving to the right from $t$ and those vertices get indices $n-1,n-2,...$etc until I end up at $s$ again. But then when I want to do the inductive step I get stuck. I start with a 2-connected graph $G$ with more than one ear and I remove the outer ear. The remaining is still a 2-connected graph and by the induction hypothesis I get some ordering. Then I am unsure of how to order the outer ear. Please help.