Well, there is an easier approach.
For each point 1 to N, run DFS from another point and check if all points are reachable from that one point. The key is that in an undirected graph, if we can reach all points from one point, then we can reach all points from all other points.
So the problem becomes:
For each point forbidden from 1 to N:
Run DFS from any point (!= forbidden) and check if all nodes other than forbidden are reachable without going through forbidden.
If yes, then it is not a critical intersection.