TITLE: The Petersen graph is 3-connected
QUESTION [6 upvotes]: This is obvious, but is there a simple/elegant way to show that the Petersen graph has no vertex cuts of size 2? 
One could just look at all possible vertex cuts of size 2 and observe that they don't disconnect the graph, but technically there are ${10 \choose 2} = 45$ possibilities. In reality most of these cuts result in the same (connected) graph. 
So I'm just asking for a clear and simple way to prove that there are no vertex cuts of size 2.

REPLY [8 votes]: Think what happens when you delete the vertices one by one.
Claim 1 If we delete one vertex from the Petersen graph, we get a Hamiltonian graph.
This follows immediately: it is irrelevant which vertex we delete, by symmetry, and once we delete one vertex it is easy to find a Hamiltonian Cycle.
Claim 2 If we delete one vertex from a Hamiltonian graph, we get a connected graph. (Actually we get a graph which has a Hamiltonian path).
Proof: The Hamiltonian cycle becomes a Hamiltonian path after erasing the vertex.
Note: In claim 2, I didn't say that the graph becomes semi-Hamiltonian because it could still be Hamitonian (for example $K_n$ has this property).