$D_i$ contains at least one edge. Moreover, each $D_i$ is bipartite. For $i \in \{1,\ldots,r\}$, we denote the two non-empty bipartition classes of $D_i$ by $D_i^1$ and $D_i^2$ such that $|V(D_i^1)|\leq |V(D_i^2)|$. The following claim is crucial.

[[[*Claim .* ]{}]{}[*\[clm:2\]For $i\in\{1,\ldots,r\}$, we must either colour all vertices of $D_i^1$ with colour $1$ and all vertices of $D_i^2$ with colour $2$, or vice versa.*]{}\
]{} We prove Claim \[clm:2\] as follows. Suppose that $D_i^1$ contains a vertex with the same colour as a vertex of $D_i^2$. As $D_i$ is connected and bipartite, this means that $D_i$ contains an edge $vw$ whose end-vertices are either both coloured $1$ or coloured $2$. In the first case, we obtain a $1$-edge. In the second case the vertices $u$, $v$ and $w$ form a $2$-cycle in $G$. Hence we must use colours $1$ and $2$ for different partition classes of $D_i$. This proves Claim \[clm:2\].

We now proceed as follows. First suppose that $S_2 \cup T_2$ is non-empty. If we coloured a vertex in $S_2$ (respectively $T_2$) with colour $1$, then every vertex in $T_1'$ (respectively $S_1'$) must be coloured $2$ and therefore every vertex in $S_1'$ (respectively $T_1'$) must be coloured $1$ by Claim 