Suppose $G = (V,E)$ is a finite and connected planar graph with a given fixed embedding in $\mathbb R^2$. Now suppose we remove all vertices that are incident to an edge belonging to the outer face of the embedding, and suppose we have a non-empty collection $V'$ of vertices left (i.e. the given embedding is not outerplanar).
Then if $U \subseteq V'$ we can find a cycle of vertices in $V \setminus V'$ such that $U$ lies in the interior region of this cycle.
How to prove this "obvious" fact?
The "obvious" fact is false. Consider the following counterexample. Let $G$ be two copies of $K_4$ embedded into a plane joined by an edge lying in the outer face, as below.
Given this embedding of $G$ and your construction in the question, the set $V'$ is given by $V'=\{v_4,v_8\}$. However there is no cycle in $G$ such that $v_4$ and $v_8$ lie on the interior of that cycle.
Suppose you make the further assumption that $G$ is $2$-connected. Then you can prove a lemma that the vertices and edges that make up the outer face in a planar embedding of $G$ form a cycle. That cycle should contain any subset of $V'$.