Prove that for any 2 edges $e_1$, $e_2$ of a connected graph $G$ there exists a path having $e_1$ as its first edge and $e_2$ as its last edge.
Attempt
Let $e_1=u_1v_1$ and $e_2=u_2v_2$ in $A(G)$. Since $G$ is connected, there exists a $v_1u_2-$path. This path can contain either $u_1$ or $v_2$, if it contains both vertices we are done. Suppose it contains $u_1$, since $v_2$ is adjacent to $v_1$ this produces a $v_1v_2-$path. Analogous the other case.
If it contains neither, since $u_1$ is adjacent to $v_1$ and $v_2$ is adjacent to $u_2$, this produces a $u_1v_2-$trajectory that has $e_1$ as its first edge and $e_2$ as its last edge.
Is it ok?