Let $G=(V,E)$ be a simple, disconnected graph with at least $2$ vertices. $G$ has exactly two vertices of even degree. Let $G'=(V,E')$ be the complement graph of $G$. Prove that $G'$ has Eulerian path.
I think we have two cases when $|V|$ is odd or even. If $|V|$ is odd then the degrees of vertices $x,y$ (whose degrees are even in $G$) will be odd in $G'$ while all the degrees of all other vertices will be even. Also $G'$ is connected (the proof is here). This satisfies the definition of Eulerian path: there's either $0$ or $2$ vertices of odd degree in a graph.
But what if $|V|$ is even? Then the degrees of $x,y$ are even and there're $|V|-2$ vertices of odd degree which is not necessarily a Eulerian path.
How can I prove this when $|V|$ is even?