I'm currently working in the following graph-theory excercise:
Let $G_1$ and $G_2$ be two Eulerian graphs with no vertex in common. Let $v_1$ $V(G_1)$ and $v_2$ $V(G_2)$. Let $G$ be the graph obtained from $G_1 \cup G2$ by adding the edge $v_1v_2$ . What can be said about $G$?
My answer is that as $G_1$ and $G_2$ are Eulerian graphs then by Euler's theorem all of their vertex are even, but by adding the edge $v_1v_2$ then $v_1$ and $v_2$ are odd, then $G$ is a semi-eulerian graph. Is that correct? Thanks in advance for any critic or correction.
Your proof seems fine.
Alternatively, you could be more direct if you wanted. You may use the new edge to get from $G_1$ to $G_2$, but you can't get back. So clearly $G$ can't be Eulerian. But you can find a path starting and ending in $v_1$ which walks all the edges of $G_1$, then you can go over the new edge, then walk all the edges of $G_2$. So $G$ is semi-Eulerian.