Proof
(related to Lemma: Dual Graph of a All Faces Contained in a Planar Hamiltonian Cycle is a Tree)
 Assume, \(G(V,E)\) is a simple, planar, and Hamiltonian graph with \(n\) vertices.
 Take any planar drawing \(\mathcal D\) of \(G\).
 Construct the corresponding dual graph \(G^\ast_{\mathcal D}\).
 Draw the Hamiltonian cycle \(C^n\) of \(G\) in \(\mathcal D\) and identify vertices \(v_1,\ldots v_m\) of \(G^\ast_{\mathcal D}\), which are enclosed by \(C_n\) in \(\mathcal D\).
 We claim that the subgraph induced in \(G^\ast_{\mathcal D}\) by those vertices, i.e. the graph \(G^\ast_{\mathcal D}[v_1,\ldots v_m]\) is always a tree. Proof by contradiction:
 Assume, \(G^\ast_{\mathcal D}[v_1,\ldots v_m]\) is not a tree.
 Then it contains at least one cycle, say \(C_k^\ast\).
 Therefore, the drawing of \(C_k^\ast\) inside the planar drawing \(D\) contains at least one face, since either \(C_k^\ast\) is a face itself, or it has some chords forming a face. Denote this face by \(f^\ast\).
 By definition of the dual graph \(G^\ast_{\mathcal D}\), there must be a vertex \(w\) of \(G\) inside the face \(f^\ast\) we find in the planar drawing of the cyclic subgraph \(G^*_{\mathcal D}[v_1,\ldots v_m]\).
 But \(w\in f^\ast\) cannot be contained in the Hamiltonian cycle \(C^n\), since \(C^n\) surrounds the hole subgraph subgraph \(G^\ast_{\mathcal D}[v_1,\ldots v_m]\), and thus is not adjacent to \(w\).
 This a contradiction to the hypothesis that \(C^n\) is a Hamiltonian cycle.
 Therefore, \(G^\ast_{\mathcal D}[v_1,\ldots v_m]\) must be a tree.
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References
Bibliography
 Piotrowski, Andreas: Own Research, 2014