This probelm is very tricky:one idea kills it Solution:Take the obvious graph interpretation: we have $\deg v\ge 2$ and every cycle doesn't have any chords. Now take the maximal path $v_1v_2\dots v_k$.By maximality all the neighbours of $v_1$ and $v_k$ are in this path. Lemma: $\deg v_1=\deg v_k=2$ Proof: by contreadiction assume $\deg v_1\ge 3$ and assume $v_i$ and $v_j$ are two neighbours $2<i<j\le k$.Then take the cycle $v_1v_2\dots v_j$ and observe it has the chord $v_1v_i$, false !Similar the other case. $\blacksquare$ Now the killer observation is that $v_i$ is the other neighbour of $v_k$ then take the path $v_1\dots v_iv_k\dots v_{i+1}$.It is another maximal path so by the lemma, $\deg v_{i+1}=2$ and so $v_k$ and $v_{i+1}$ have a common neighbour and degree $2$.The end $\blacksquare$.

I think that the official solution was something like this. Another possible approach was presented here.