Is this graph connected? Let $G$ be a graph of size at least $2$ such that any two edges are on a cycle. 
Is there something we can say about the connectivity of $G$? 
It seems like this condition doesn't imply connectivity because I can have a graph that is a triangle plus an isolated vertex, which seems to satisfy the hypothesis.
Am I missing something?
Context: An exercise in Modern Graph Theory by Béla Bollobás asks to prove a similar statement.
 A: You're right that isolated vertices cause problems. The more interesting statement is that if $G$ has no isolated vertices, then this property implies that $G$ is $2$-connected. In fact, the two statements are equivalent: $G$ is $2$-connected if and only if any two edges of $G$ lie on a common cycle and $G$ has at least two edges and no isolated vertices.
To get $2$-connectivity out of this condition, let $x,y$ be any two non-adjacent vertices of $G$. Pick two edges, one out of $x$ and one out of $y$, and find a cycle through them. This is a cycle containing $x$ and $y$, giving us two vertex-disjoint $x,y$-paths. Now we see that no vertex we delete in $G$ can disconnect $x$ and $y$: deleting a single vertex can destroy at most one of these two $x,y$-paths. Since this is true for any $x,y$ (if $x$ and $y$ are adjacent, no vertex deletion can disconnect them), $G$ must be $2$-connected.
(If $x$ or $y$ were an isolated vertex, this proof would fail at the first step, where we pick an edge out of $x$ and an edge out of $y$.)
We should also separately check $K_n$ (whose connectivity is a special case, defined to be $n-1$). It is $2$-connected for $n \ge 3$, but also vacuously has the cycle property for $n=2$, so we should ask for our graph to have at least $2$ edges, to eliminate the $K_2$ case.
A: Your counterexample is correct.  But we can ask a similar question: consider a graph $G$ with no isolated vertices, and has the property that any $2$ edges $e_1$ and $e_2$ are on a cycle (in other words, there exists a [connected] path from any vertex of $e_1$ to any vertex of $e_2$, since they are in a cycle containing both edges)
But assume that $G$ is not connected, so it has at least 2 connected components $G_1$ and $G_2$.  So we can pick an edge $e_1$ from $G_1$ and an edge $e_2$ from $G_2$.  Clearly there is no cycle containing both edges since the components are disconnected.  So such a graph $G$ must be connected
