Let $G = (V,E)$ be a planar connected graph with a fixed embedding. Then if $G$ is not a tree, how to prove that the boundary of every face contains at least one cycle?
Seems to me quite difficult to find the right "topological" arguments in $\mathbb R^2$. Intuitively, if the boundary of some face does not contain a cycle, then this face is not "separated" from other faces, hence we just have one face, and our connected graph must be a tree. But this kind of argument is to informell, how to make it formal by arguing with the embedding, properties of $\mathbb R^2$ (Jordan curves and so on...)?
EDIT: Regarding the comment of looking at a sequence when traversing the edges (or vertices) in a specific order, that would not give me necessarily a cycle, see the picture below where this sequence ends (or when I go on the first repetition would not give a cycle).
