Here is a quick reduction.  Hopefully someone else can finish it off.  Since $G$ is 2-connected, it has an ear-decomposition starting with the cycle $C$.  Next, when building the ear-decomposition, for as long as possible always choose ears $P$ such that both ends of $P$ are in $C$ and $P$ has two edges.  Now consider the last ear $P'$.  If $P'$ is just an edge $e$, then $G \setminus e$ is 2-connected and we win by induction.  Thus, $P'$ has at least two edges.  If $P'$ has at least 3 edges, then let $G'$ be the graph obtained from $G$ by replacing $P'$ by a path of length 2.  Note that $G'$ is 2-connected, and every vertex in $C$ still has a neighbour outside of $C$ in $G'$.  Thus, by induction, $G'$ and hence $G$ has a cycle longer than $C$.  Thus, $P'$ has exactly two edges.  If at least one end of $P'$ is not in $C$, then by replacing $P'$ with a single edge, we win by induction.  Thus, both ends of $P'$ are in $C$.  Therefore, *every* ear is a $C$-path with two edges. We can thus colour the edges of $C$ red and replace each ear with a blue edge.  We now have a graph $G''$ with $V(G'')=V(C)$, and where every vertex is incident to a blue edge.  I think it should be easy to show that such a graph has a cycle longer than $C$ (red edges have length 1, and blue edges have length 2).