Theorem. Let $G$ be a plane graph with at least 3 edges drawn on $\mathbb{R}^2$. Then every face of $G$ is bounded by at least 3 edges.
We define vertices to be points in $\mathbb{R}^2$ and an edge is an arc. An arc is an injective continuous function $f : [0,1] \to \mathbb{R}^2$ that is a union of a finite number of straight line segments. We define a face of $G$ to be a maximal arcwise connected subset of $\mathbb{R}^2 - E(G)$. The boundary of a face $F$ is the set of edges whose interior meet $F$.
I am using Diestel's theorems mainly. It is proven that for 2-connected plane graphs every face is bounded by a cycle, so the theorem holds if $G$ is 2-connected.
Is it easy to (rigorously) show the above theorem in general?
Let $f$ be a face of $G$, and let $H$ be its boundary. If $H$ contains a cycle, we are done. Otherwise, $H$ is a plane forest, so $H$ by itself only has one face $f'$ by Proposition 4.2.3. So $f \subseteq f'$. If there is a point $x \in f' - f$, consider an arc in $f'$ from $y \in f$ to $x$. The boundary of $f$ is $H$, so the arc meets $H$. But the arc is in $f'$ so it should not meet $H$, contradiction. Thus $f = f'$. But now $H \cup f = \mathbb{R}^2$, so that $H = G$. Since $G$ has at least 3 edges, so $H$ has at least 3 edges.