Given a graph G, if X(G) = k, and G is not complete, must we have a k-colouring with two vertices distance 2 that have the same colour? As the title asks, If given a graph, $G$, with chromatic number $k$, and $G$ is not complete must there exist a $k$-colouring of $G$, $f$, where there are two vertices $x,y$ such that $d(x,y) =2$ and $f(x) = f(y)$. 
Note that not all $k$-colourings of $G$ have this property. Just consider the $9$-cycle with $k=3$ and colour it $a-b-c-a-b-c-a-b-c$, but the $9$-cycle has a $3$-colouring $a-b-a-b-a-b-a-b-c$ which does. 
In the case where $k =3$, start by noticing that $G$ has a cycle C, where $|V(C)|>=3$. Furthermore you can find a chordless cycle with this property. 
If possible, identify vertices $v,w$ on $C$ which are distance $2$ apart which have the same colour and repeat until no longer possible. Now given any three adjacent vertices on the cycle, $x,y,z$ where the edges are $xy$, $yz$, if there is a vertex $p$ adjacent to $y$, the colour of $p$ is either the same as $x$ or the same as $z$ and then we identify whichever happens to work. Continue this procedure until you are left with just a cycle, at which point you can unidentify vertices and give them the same colour as which they originated. The last two vertices which are unidentified are the desired vertices. 
Thanks for any help :)
Edit: $G$ is a simple connected graph

I'll give an alternative proof to Leen's.
Suppose we have a graph $G$ where $\chi(G) =k$ and $G$ is not complete. We will proceed by showing that using kempe chains we can always get two vertices distance $2$ with the same colour.
Now take any $k$-colouring of $G$, $\alpha$. Since $G$ is not complete, there is at least one colour class $C$ of $G$ which has more than one vertex in it. If there are two vertices of $C$ which are distance $2$ apart, then we are done. Otherwise let $x,y \in V(C)$. Since $G$ is connected, there is a shortest path, $P$ between $x$ and $y$.  Then there is a vertex $z \in N(x)$ which lies on $P$. Apply a kempe chain operation on the $x$ and $z$. This results in a new colouring $\alpha'$, and now we have reduced the distance between colours in $C$, as $d(x,y) = d(y,z) +1$. Now we can apply this same technique on $z$ and $y$, continually reducing the distance until we have a $k$-colouring where there are two vertices which have the same colour and are distance $2$ apart.
