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 $$C \ne G$$ 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.

Therefore 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 $$C$$

Thus it has a degree of at least $$3$$.

But $$G$$ is a cycle and every vertex in a cycle has degree $$2$$: a contradiction.