Prove that a $(2k+2)$-edge colouring of a $(2k+1)$-regular Class two graph, $G$, has at least two edges of each colour. Let the $2k+2$ colours belong to the set $\{1,2,...,2k+2\}$. I thought the best way to answer this question would be to proceed by contradiction, and assume that $G$ has an edge with some colour, $1$, that only appears once.
Let $uv$ be the edge with colour $1$. Then every vertex of $G$ that is not $u$ or $v$ is incident with an edge of every colour except 1. Since $G$ is $2k+1$-regular, each of $u$ and $v$ has $2k$ other edges incident to it, and since we can't use $1$ again, the colours for these remaining $2k$ edges must be chosen from the remaining $2k+1$ colours.
Of course if $u$ and $v$ both have the same $2k$ colours for their remaining edges, say from the set $\{2,...,2k+1\}$ then we just re-colour $uv$ to be the unused colour, $2k+2$ and obtain a $(2k+1)$-colouring of $G$, which is a contradiction since G is Class 2.
However, I'm unsure what to do in the case where the remaining $2k$ colours are not the same between $u$ and $v$. Obviously since there are only $2k+1$ colours to choose from, $u$ and $v$ will only differ by one colour in this case, but I don't know how to proceed and get a contradiction, nothing I've attempted so far has worked.
Any help would be greatly appreciated.   
Edit: I've posted a potential solution that I came up with and would appreciate if somebody could verify it works.
 A: Okay so I've thought some more about it and I think I have a potential contradiction for the second case, but I would appreciate if somebody could verify this works.
Let $G$ have order $n$. Assume, without loss of generality, that the colours incident at u are $\{1,2,4,...,2k+2\}$ (ie: all the colours except 3), and that the colours incident at v are $\{{1,3,4,...,2k+2}\}$ (ie: all the colours except 2). Since every other vertex of $G$ is incident with an edge of every colour except $1$, this means there are $n-1$ vertices incident with an edge of colour 2. But then these edges form a 1-factor in $G-v$. But $G-v$ has odd order, since G has even order (because it's $2k+1$ regular), therefore $G-v$ cannot have a 1-factor. Contradiction? 
A: A partial answer: Since the graph is $(2k+1)$-regular, then it has at least $2k+2$ vertices. Since the only $(2k+1)$-regular graph on $2k+2$ vertices is $K_{2k+2}$, which is Class 1, our graph must have at least $2k+3$ vertices. 
Select any subset of $2k+3$ vertices and observe that edges incident with each vertex is missing exactly one color from the $2k+2$ color set, thus there must be some pair of vertices from our subset that have the exact same color assignments on the incident edges. So we know all but one of our colors is duplicated at least twice.
I’m not sure how to verify that the missing color appears twice elsewhere though.
