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
 A: Let $G$ be a connected graph with $\chi(G)=k$ and no $k$-coloring such that two vertices at distance 2 have the same color.
Let $c$ be an arbitrary (proper) $k$-coloring of $G$ and let $v$ be an arbitrary vertex of $G$ with color, say, red.
Let $B_i$ be the set of vertices of $G$ that have distance $i$ from $v$.
Note that $B_0\cup B_1=N[v]$: the closed neighbourhood of $v$.
Claim 1: Every color occurs exactly once on $N[v]$.
Claim 2: There is no vertex outside $N[v]$.
Claim 3: $G$ is complete.
The last claim obviously implies that the answer to your question is "Yes".
Proof of claim 1:
Suppose blue is a color that does not occur on $N[v]$.
Let $i$ be the smallest index such that blue occurs in $B_i$ and $w$ a vertex in $B_i$ with color blue.
Note that $i\geq2$.
A shortest $v,w$-path has length $i$ and hits $B_{i-2}$ in exactly one vertex $x$.
Since blue does not occur in $B_j$ for $j\leq i-1$ we can recolor $x$ with blue and we found a forbidden coloring. Contradiction.
Clearly red cannot occur twice on $N[v]$.
Suppose blue is a color that occurs (at least) twice on $N[v]$, say on $x$ and $y$.
$x$ and $y$ cannot be adjacent, since $c$ was a proper coloring. But then they have distance 2 and the same color. Contradiction.
Proof of claim 2:
If there is a vertex outside $N[v]$ then there certainly must be a vertex $w$ in $B_2$.
This vertex cannot be red, or we have two red vertices at distance 2.
Suppose $w$ is blue.
Let $x$ be the blue neighbour of $v$ (that exists because of claim 1).
Now $x$ has no red neighbour except $v$ and certainly no blue neighbour
and $v$ has no blue neighbour except $x$ and certainly no red neighbour.
So we can swap the colors of $x$ and $v$ and now we have a proper coloring with two blue vertices ($v$ and $w$) at distance 2. Contradiction.
Proof of claim 3:
Claim 1 and 2 together already show that $G$ has exactly $k$ vertices and each color occurs exactly once.
So if two vertices are not adjacent we can recolor one of them and obtain a proper coloring with two vertices of the same color. Contradiction.
(Since $v$ was an arbitrary vertex, claim 3 also can be proved with a symmetry argument).
A: 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.
