Is every connected, finite, undirected graph in which all vertices are degree two a cycle? 
Is every connected, finite, undirected graph in which all vertices are degree two a cycle?

I feel like this is true, and this is my attempt at a proof:

Start at any vertex. From this vertex, leave to an adjacent vertex, and from there leave to the next and so on. Repeat this process, and every time you get to a vertex, you'll have exhausted one of its outgoing edges, so you will only have one choice -- to move on to the next vertex. Since the graph is finite and connected, eventually you will exhaust all unique vertices and since there is still one vertex (the first vertex we start at) with one unused edge, we must return to it.

 A: No. You have not used the assumption that the graph is connected, nor the assumption that the degree of each vertex is exactly two. Your argument is the argument one would give to prove that a finite graph, in which every vertex has degree at least $2,$ contains a cycle. In order to prove that the graph is a cycle, you have to prove that it has no more vertices or edges besides the vertices and edges of your cycle. You can prove that using the assumptions that the graph is connected and that the vertex degrees are exactly $2.$
A: Yes, your proof looks good to me.
A: Let $X=(V(X),E(X))$ be the given graph.
We know that $X$ is 2-Regular,connected,simple, and undirected.
Aim : Want to show $X$ is Cycle.
Proof : Let $|V(X)|=n,$ and $V(X)=\{v_1,v_2,\dots,v_n\}$.
Wlog let us assume that $\{v_1,v_2\}$ and $\{v_1,v_n\}\  \in E(X)$ ,
consider $v_2$ since $deg(v_2)=2$, $v_2$ has to be in adjacent with any $v_i \in V(X)$, suppose $\{v_{2},v_n\}\in  E(X)$ this will contradict the fact that graph $X$ is connected ! why ? {hint Look at the vertex $v_3$,it has got no path with $v_1$ which is a contradiction to the fact that $X$ is connected.}
So $\{v_{2},v_n\}\not\in  E(X)$ again one can assume $\{v_{3},v_2\}\in E(X)$ again similar argument we put on $v_n$ and $v_2$ will work here and will get that $\{v_n,v_3\}\not\in E(X)$. 
Repeat this process until you hit $v_{n-1}$ and there you can say that $\{v_{v-1},v_n\}\in E(X)$ which will result in saying that $X$ is cycle.
Basically this is your idea of proof only, I just written elaborately so that it will be clear to everyone why this fact is true.The numbering I have given to $V(X)$ is just random only so it doesn't affect our proof. 
