Let $G$ be a (simple) graph and let $a,b\in V(G)$. If every proper (vertex) colouring $c$ of $G$ has $c(a) \neq c(b)$ then for any proper coloring $c_0$ of $G$, there is an $a$-$b$-path, $P=av_1v_2...v_nb$ with $c_0(v_i)=c(b)$ for $i$ odd and $c_0(v_i)=c(a)$, otherwise.
Proof: Let $c$ be a proper colouring of $G$. Define $V_{ij}=\{u\in V(G) : c(u)=c(a) \text{ or } c(u)=c(b)\}$ and consider $G_{ij}$ to be the subgraph of $G$ induced by $V_{ij}$. Let $C$ be the component of $G_{ij}$ that contains $a$. Note that we can switch colours $c(a) \leftrightarrow c(b)$ in $C$ and create a new proper colouring of $G$. Otherwise, if there is some conflict with some $v\in V(G)\setminus V(C)$ that would necessitate $v$ to have either of the two colours $c(a)$ and $c(b)$, in which case it would be in $V(C)$. Thus, if $b$ is not in $V(C)$ then we could have a proper colouring $c_0$ with $c_0(a)=c_0(b)$. Thus $b$ must be in $C$, in which case there is an $a$-$b$-path. Then if the path does not have alternating $c(a), c(b)$ colours, it must have some third colour. But then again we could have some proper colouring $c_1$ with $c_1(a)=c_1(b)$. \qed
Does this statement and proof make sense? Is there some easier way to see this fact or a generalization of this fact? This was taken from a proof of the 5-colour-theorem where the fact was stated unproven.