My attempt at a proof is:
Let $G$ be a $2$-connected graph. Since $G$ is $2$-connected, $G$ contains no bridges - otherwise, we could pick an arbitrary bridge, remove one of its ends, and render $G$ disconnected - this contradicts $G$'s $2$-connectivity. Recall that an edge is a bridge, if and only if it does not lie on a cycle. Since $G$ contains no bridges, all of its edges must lie on some cycle, and hence, $G$ contains at least one cycle.
Is this the correct idea? If not, any suggestions?
It looks good, but note that removing one end of a bridge only disconnects the graph if there are other vertices on that side of the bridge. (This is ok because there must be other vertices on at least one side.)