Prove that in 2-edge-connected graph, each vertex lies on a cycle.

Let $G=(V,E)$ graph, $|V| \geq 3$. Suppose that there exists an edge $e=(u,v) \in E$ such that $e$ does not lie on a cycle. Then every path from $u$ to $v$ must contain $e$. If either $u$ or $v$ are a leaf, then removing the vertex that connects the leaf from the rest of the graph makes the leaf disconnected. Otherwise, define the following partition of $G$ without $e$: \[ G_u=(V_u,E_u), \quad G_v=(V_v,E_v) \] where \[ V_u = \{w \in V \mid \text{ there is a path from $w$ to $u$ that does not contain $e$} \} \] and $E_u = \{(w,t) \in E \mid w,t \in V_u\}$ and $V_v, E_v$ are defined analogously. It is immediate to see that these sets are a partition of $(V,  E \setminus \{e\})$. Now, if we remove $e$, then there is no path to go from any $w \in G_u$ to any $t \in G_v$, so the graph becomes disconnected, a contradiction because it is $2$-connected.