Cycle does not Contain Subcycles

Theorem
If a graph $$G$$ is a cycle, then the only cycle that is a subgraph of $$G$$ is $$G$$ itself.

Proof
Suppose that $$G$$ contains a subgraph $$C$$ that is a cycle and $$C\neq G$$ is non-empty. Then there is some vertex $$v$$ that is not in $$C$$. Let $$u$$ be any vertex of $$C$$. Since $$G$$ is a cycle, it is connected, so there is a walk from $$u$$ to $$v$$ in $$G$$. There must be some vertex $$x$$ that is the last vertex in $$C$$ along that walk. Therefore, $$x$$ is adjacent to a vertex not in $$C$$, giving it a degree of at least 3. But $$G$$ is a cycle and every vertex in a cycle has degree 2: a contradiction.