For a 3-connected counterexample, you can also take a 'truncated' Peterson graph, truncating meaning that we remplace each vertex of $Pete$ by a small triangle (think of a truncated cube). Then a cycle $G[V_1]$ must hit each triangle and thus alternate between triangle edges and edges of the original $Pete$. The latter edges would form a Hamilton cycle of $Pete$, which doesn't exist.

Note that similar truncating constructions for higher connectivity $k>3$ won't work, as for $k=4$ a non Hamiltonian 4-regular 4-connected graph just doesn't exist (if my memory serves), and for $k>4$ a 'truncated vertex' $C_k$ might be visited more than once.