misread problem

EDIT: Didn't realize non-leaves could have back edges Anyway the proof should be correct now. Take the obvious graph $G$. WLOG assume $G$ is connected, else look at a connected component. The given condition implies that every cycle is chordless. Do DFS from a random vertex to get a spanning tree $T$. Note that every back edge in a DFS is from a vertex to an ancestor of the vertex. By the chordless condition, every vertex has at most one back-edge starting at that vertex, and since every degree is at least $2$, every leaf has exactly one back-edge, i.e., every leaf has degree exactly two. For any $u_1,u_2$, let $P(u_1,u_2)$ denote the unique path from $u_1$ to $u_2$ in $T$. For any leaf $l$, let $f(l)$ denote the other endpoint of the back-edge at $l$, and let $g(l)$ be the first vertex after $f(l)$ in the path from $f(l)$ to $l$ in $T$. Clearly $g(l) \neq l$. Choose a leaf $l$ such that $f(l)$ is furthest away from the root. We claim that $g(l)$ has degree $2$. We will then be done, as $l$ and $g(l)$ have a common neighbor $f(l)$. Since $g(l)$ already has two edges: One going to $f(l)$, and one going along the path from $g(l)$ to $l$. We have to show that there are no other edges. Let $C$ be the cycle consisting of the path in $T$ from $f(l)$ to $l$, and the back-edge from $l$ to $f(l)$. We first show that degree of $g(l)$ in $T$ is $2$, i.e., there are no other branches at $g(l)$. Assume the contrary; then there is a leaf $l' \neq l$ that is a descendent of $g(l)$ along the other branch. $f(l')$ is closer to the root than $f(l)$, by our given condition. Then $f(l)g(l)$ is a chord in the cycle $P(f(l'),f(l))f(l)lP(l,g(l))P(g(l),l')l'f(l')$, contradiction! Now we analyze back-edges at $g(l)$. First, suppose there is a back-edge starting at $g(l)$. The other end, $u$, must be an ancestor of $f(l)$. But then, $f(l)g(l)$ is a chord in the cycle $g(l)uP(u,f(l))f(l)lP(l,g(l))$, contradiction! Finally, suppose some back-edge $ug(l)$ ends at $g(l)$. $u$ cannot be in $C$ as $C$ is chordless, so $u$ cannot be an ancestor of $l$. $u$ cannot be a leaf, as then $f(u)=g(l)$ is further away from the root than $f(l)$. Hence $u$ has a leaf descendent, say $l'$. Note that $f(l)$ is an ancestor of $l'$, so again by the condition on $l$, $f(l')$ must be an ancestor of $f(l)$, which is an ancestor of $g(l)$. But then, $g(l)u$ is a chord in the cycle $l'f(l')P(f(l'),l')$, contradiction! Hence we get a contradiction in every case, so degree of $g(l)$ is $2$, as required.

https://artofproblemsolving.com/community/c6h3313173p30606863