I just hate having to come here to ask questions. It's like accepting defeat that you can't solve it yourself.. but I've been trying to solve this for hours and can't find the intuition to solve the problem..
Let 𝐺 be a 4-critical graph having a vertex cut 𝑆 of cardinality 4. Prove that the induced subgraph 𝐺[𝑆] has at most four edges.
Clearly $G[S]$ can't have 6 edges, since then it would be the complete graph on 4 vertices and thus 4-chromatic $\implies$ $G$ not 4-critical. I've been able to show existence of 4-critical graphs that have a vertex cut of cardinality 4 such that the graph induced by the vertex cut has 4 edges. I fail to see why something interesting happens with 5 edges. I've been trying to consider something like: showing that if there are 5 edges, then there exists a 3 colouring of the S-lobes (plural since S is vertex cut, 3 colouring since G is 4-critical) such that they are consistent on S $\implies$ G is 3-colourable.