I am looking for an error in the proof but I am not certain about it. Pretty sure it has something to do with how there is not always a cycle of length 3.
Theorem 1. For every (undirected) graph $G = (V; E)$ without loops, if $|V| \ge 3$ and for every vertex has a degree of at least $2$, there exists a cycle of length $3$.
Proof. We proceed by induction on $|V|$. As a base case, consider $|V| = 3$. For any graph on three vertices, if all vertices have degree $2$, then every vertex is connected to both other vertices. Thus there is always a cycle of length $3$. To prove the inductive step, let $G$ be a graph with $k$ vertices, and construct a new graph $G_0$ on $k + 1$ vertices by adding one new vertex to $G$ and $2$ edges incident to the new vertex. Since $G$ has $k$ vertices, by the inductive hypothesis it has a cycle of length $3$. Thus the graph $G_0$ contains the same cycle of length $3$. This completes the proof.