The Petersen graph is 3-connected 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.  
 A: 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).
A: We note that Petersen's graph consists of two disjoint circuits of length $5$ and a perfect matching between these two groups of $5$ vertices. If we delete two vertices in one of these two circuits, it will definitely remain connected, since we still have a circuits of length $5$ and matching edges which will connect the remaining vertices in another circuit. If we delete one vertex from each of these two circuits, then it is still connected, since they become two paths of length $4$ and there is still some matching edge connecting them.
