Two cycles covering vertices I consider an edge-coloured graph with the colours red and blue. Gyarfas proofed in his paper http://www.renyi.hu/~gyarfas/Cikkek/16_Gyarfas_VertexCoveringsByMonochromaticPathsAndCycles.pdf the existence of two cycles covering the vertices and intersecting on at most one vertex. 
He considered the longest path consisting of a red path followed by blue path (such a path $P$ is Hamiltonian). If a vertex $v$ is not covered it must be joined in blue to the origin of $P$ and in red to the end of $P$. Then you can cover the vertices of $P$ and $v$ using the edge from the starting point of $P$ to the end point (lets call them $a$ and $b$). Therefore there exists a monochromatic cycle $C$ and a monochromatic path $P$ with different colors partitioning the vertex set. 
I am asking for help with the drawing. I think it looks like  but where is the cycle $C$?
 A: What Bessy and Thomasse are ultimately arguing in their second paragraph is the following:

Lemma. Let $n\geq 2$. For every colouring of $E(K_n)$ into two colours, there exists a monochromatic cycle $C$ of size at least two and  monochromatic path $P$, coloured differently from $C$, such that  $P$ and $C$ are vertex-disjoint and $V(P)\cup V(C)=V(K_n)$.

It should be noted that they are considering empty sets, singleton vertices and edges as cycles, and empty sets as paths.
They justify the lemma in three steps, and I think that perhaps your problem is that you're confusing the proof of the above with the proof of the first of these steps. Anyway, here is a more detailed proof.
Step 1: There exists a Hamiltonian path $H$ such that $H$ consists of a red path followed by a blue path.  
Proof: Call any path consisting of a red path followed by a blue path a "red-then-blue" path. Then we show that any red-then-blue path that is not a Hamiltonian path can always be extended to a longer red-then-blue path.
To do this, suppose $Q$ is a red-then-blue path, say from $a$ to $b$ but that there is a vertex $v$ not on $Q$. Something like the following picture (the same as yours but drawn a bit differently. $Q$ runs along the top here):

(Note that there is no $C$ in this picture. That part comes later.)
Now we consider the edge $ab$. If it is red, then we get a longer red-then-blue path as outlined in green here:

If $ab$ is blue, then we get a longer red-then-blue path as outlined in green here:

This completes the proof of Step 1.
Step 2: There exists a Hamiltonian cycle consisting of a single red path and a blue path.
Proof: Take a red-then-blue Hamiltonian path $Q$, say from $x$ to $y$, then since $xy$ is either red or blue, we must have that $Q\cup xy$ is a red-then-blue Hamiltonian cycle. This completes the proof of Step 2.
Proof of Lemma: Let $H=(v_1,\ldots,v_n)$ be a Hamiltonian cycle consisting of a single red path, say $(v_1,v_2,\ldots,v_k)$, and a blue path $(v_k,v_{k+1},\ldots,v_n,v_1)$.
If $H$ is all in one colour, then we take $C=H$ and $P=\emptyset$. Note that this also takes care of the case $n=2$, so we can assume $n\geq 3$ and that $H$ has at least two colours.
Consider the edge $e$ between $v_1$ and $v_k$. If this edge is red, then take $C$ to be the red cycle $(v_1,v_2,\ldots,v_k,v_1)$, and take $P$ to be the blue path $(v_{k+1},v_{k+2},\ldots, v_n)$. If $e$ is blue, take $C$ to be the blue cycle $( v_1,v_k,v_{k+1},\ldots,v_n,v_1)$ and $P$ the red path $(v_2,\ldots,v_{k-1})$. In either case, $C$ contains $e$ and so has length at least $2$.
