Let $F$ and $H$ be 2 disjoint non-Eulerian regular graphs and let $G=(F+H) \bigvee K_1$. Prove that $G$ is Eulerian
Here is what I got so far.
Let $F$ be a $r-regular$ graph of order $x$ and $H$ be a $k-regular$ graph of order $y$. Then $rx$ must be even (at least $r$ or $x$ is even) for $0\leq r\leq x-1$ and $ky$ must be even (at least $k$ or $y$ is even) for $0\leq k\leq y-1$. Let $v \in K_1 $, then $G$ is obtain by joining $v$ to every vertex in $F$ and $H$. I want to show that every vertex in $G$ are even vertices. Should I do this by case? if so I will have $v$ is an odd vertex if $x$ and $y$ are not both odd or both even.
Clearly $G$ is connected. We only need to show that all nodes in $G$ has even degree. Since $F$ and $H$ is non-Eulerian, it follows that it must have at least one vertex with odd degree. Since they are regular, it follows that all vertices in $F$ and $G$ must have odd degree. Since they are disjoint, all nodes in $F+H$ must have odd degree. Thus, all nodes in $G$ except for $K_1$ must have even degree. Since the sum of all degrees in a graph must be even, it follows that the degree of $K_1$ is also even.