Showing there is a node in the graph with only one edge
I think there'll be no different. You mean that $v_2\in N(v_1)$ and $v_1\in N(v_2)$ is allowable. While $v_1\notin N(v_1),$ $v_2\notin N(v_2),$ $N(v_1)\cap N(v_2)=\varnothing$ is still right.
Showing there is a node in the graph with only one edge
I think there'll be no different. You mean that $v_2\in N(v_1)$ and $v_1\in N(v_2)$ is allowable. While $v_1\notin N(v_1),$ $v_2\notin N(v_2),$ $N(v_1)\cap N(v_2)=\varnothing$ is still right.