Writing basic proofs about cycles? These are extremely straightforward statements, but I'm getting flustered by how someone would go about constructing proofs to solve these... 
(a) Every cycle is connected
(b) Every cycle is 2-regular
(c) Conversely, every connected, 2-regular graph must be a cycle. 
The logic behind these is very straightforward, but constructing a proof is giving me difficulty. If I could get any help, it would be appreciated! 
 A: The first two are really just a matter of having a clear definition of a cycle and working with it.
I define an $n$-cycle to be a graph with vertices $v_0,v_1,\ldots,v_{n-1}$ and edges $\{v_k,v_{k\oplus 1}\}$ for $k=0,\ldots,n-1$, where for $r,s\in\{0,\ldots,n-1\}$ we define $k\oplus\ell=(k+\ell)\bmod n$. Similarly, we define $k\ominus\ell=(k-\ell)\bmod n$. (This is just a clever trick to avoid having to treat the edge $\{v_{n-1},v_0\}$ as a special case.) Let $C$ be an $n$-cycle.


*

*If $k,\ell\in\{0,\ldots,n-1\}$, there is a unique $r\in\{1,\ldots,n-1\}$ such that $\ell=k\oplus r$. Then the sequence $\langle v_k,v_{k\oplus 1},\ldots,v_{k\oplus r}\rangle$ is a path in $C$ from $v_k$ to $v_\ell$, so $C$ is connected.

*For any $k\in\{0,\ldots,n-1\}$, the vertices of $C$ connected to $v_k$ by edges are $v_{k\ominus 1}$ and $v_{k\oplus 1}$, so $C$ is $2$-regular.
It’s the third one that requires more than just working directly with the definition of a cycle. Suppose that $G$ is $2$-regular and has $n$ vertices. Let $v_0$ be any vertex of $G$, and let $v_1$ be one of the two vertices adjacent to $v_0$. Since $\deg v_1=2$, there must be some vertex $v_2\ne v_0$ adjacent to $v_1$. In general, given adjacent vertices $v_{k-1}$ and $v_k$, let $v_{k+1}$ be the unique vertex adjacent to $v_k$ and different from $v_{k-1}$. Since $G$ has only $n$ vertices, for some smallest $k\le n-1$ we must have $v_{k+1}=v_\ell$ for some $\ell\in\{0,\ldots,k-2\}$.


*

*Show that $\ell$ must be $0$.  

*Conclude that the subgraph of $G$ induced by $\{v_0,\ldots,v_k\}$ is a $(k+1)$-cycle and a component of $G$.  

*Conclude further that if $G$ is connected, then $k+1=n$, and this cycle is all of $G$.

A: I suppose that we know the fact that for graph $G$ with minimum degree $\delta(G)\geq 2$, it must contain some cycle $C$. Now, suppose there is vertex $v\in V(G)\setminus V(C)$, then $v$ must be disconneced from $C$, otherwise some vertex in $V(C)$ must have degree at least 3. With $V(G)=V(C)$ and the $2$-regularity, we conclude that $G=C$, as all edges must be in $E(C)$.
