Proof

Let \(G(V,E)\) be a simple graph \(G(V,E)\).

It is obvious, that if \(G\) is disconnected, then it is not Hamiltonian, because it otherwise does not contain a Hamiltonian cycle.
If \(G\) is only connected, but not at least biconnected, then it contains by definition cutvertices.
In this case, it cannot contain a Hamiltonian cycle. Otherwise, every closed walk, which contains every vertex of \(G\), passes a cutvertex at least twice.
By contraposition, if \(G\) is Hamiltonian, then it must be at least biconnected.
∎

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