Every $2$-coloring of $K_n$ contains a hamiltonian ... 
Every $2$-coloring of $K_n$ contains a hamiltonian cycle that is made out of two monochromatic paths.

If there is a monochromatic hamiltonian cycle then we are done. Elseway I thought about using Dirac's theorem somehow (I know that there is in both monochromatic induced subgraphs there is a vertex with degree higher than $n/2$).


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*$K_n$ is the complete graph on $n$ vertices.

*A Hamiltonian cycle is a cycle that visits each vertex exactly once. 

*A $2$-coloring of the graph is a partition of the set of edges in two sets (for example the "blue" edges and the "red" edges).


Choose any vertex $v \in K_n$.  Then $K_n \setminus v$ is a 2-colored copy of $K_{n-1}$, and thus by induction contains a Hamiltonian cycle that is the union of a red path and a blue path.  Let $u$ be  a vertex where these two paths meet, and let $r$ be $u$'s neighbor along the red path, and $b$ be $u$'s neighbor along the blue path.
Suppose $uv$ is colored blue.  Then throw away $ur$ and instead use $uv$ and $vr$ in your Hamiltonian cycle (it does not matter what color $vr$ is).  If $uv$ is colored red, then throw away $ub$ and use $uv$ and $vb$ in your Hamiltonian cycle.  Either way, you've extended your Hamiltonian cycle to include $v$ and it is still the union of a blue and red path.
I've ignored the case where the Hamiltonian cycle in $K_{n-1}$ is monochromatic, but it is trivial to extend the cycle to $v$ in this case.
