When can we find cycles in a graph? Induction on $|V|$! EDIT: I know the other proofs, I am just looking to verify this one, Thank you!
Question: Prove that if in a graph (or multigraph), $G$, each vertex has degree at least 2, then there is a closed path (cycle) in $G$.
Solution (?): First off, we are talking about undirected graphs only. Now if there is a loop on vertex $v$, then $v-v$ is a closed path. Similarly, if there are multiple edges between two vertices, say $v_i , v_j$, then $v_i-v_j-v_i$ is a closed path. So now, assume $G$ is simple.
Let $G=(V,E)$. We will use strong induction on $|V|$. For $|V| =3$, it is trivial (we just have a triangle). Now assume the hypothesis holds for $|V|=3,4,5, \cdots ,n-1$. We need to prove it for $|V|=n$.
Now if $\exists \  v \in V$ such that $deg(v)=2$, say $v$ is adjacent to $v_1,v_2$, we can just delete $v$ and it's edges and join $v_1$ and $v_2$. And now in this new graph, clearly each vertex has degree at least $2$, so this has a cycle, so $G$ has a cycle also. (see fig. below) 
Now, otherwise, let $v$ be the vertex in $G$ with minimum degree, say $d$. Now, first randomly delete $d-2$ edges of v, and then say $v$ is adjacent to $v_1,v_2$, so delete $v$ and it's edges and join $v_1,v_2$. Now, in this graph, each vertex has degree at least $2$ also (as degrees of $v_1$ and $v_2$ are unchanged, and for all other vertex, there degree is just reduced by $1$, and since there is no vertex with $deg(v_i)=2$, each vertex still has degree at least 2). Now, this is a graph with $n-1$ vertices in which $deg(v_i) \ge 2 \ \forall \ i$ so this has a cycle and hence G has a cycle.
I have just started studying Graph Theory (and I love it :D) and I came across this problem. I think my solution is correct but I could not find it anywhere else; not even in this MSE thread, so I am now suspicious. I am especially not sure of the induction part and specifically, I am not sure how to frame the delete vertex with degree $2$ part more rigorously.
So basically, I just want to see if my proof is correct, and if so, how can I frame it more rigorously.
Thank you!
 A: This seems to end up being an awfully convoluted way to approach, as contracting a vertex $v$ and seeing what happens can be tricky if $v$ has degree 3 or larger. This is how I would approach.
First note the following: For any $k \ge 3$ let $P_k=v_1v_2\ldots v_k$ be any path on $k$ vertices in $G$. From $P_k$ we find either a path $P_{k+1}$ on $k+1$ vertices or a cycle. How? Let us first write $P_k=v_1\ldots v_k$, where $v_1$ and $v_k$ are the endpoints of $P_k$, and where $v_j$ is adjacent in $P_k$ to $v_{j-1}$ and $v_{j+1}$ for $j=2,\ldots, k-1$. Now let $w$ be a neighbor in $G$ of an endpoint $v_k$ of $P_k$ such that $w$ is not $v_{k-1}$, where $v_{k-1}$ is the vertex that is adjacent to $v_k$ in $P_k$. As $v_k$ has at least 2 neighbors there exists such a vertex $w$. If $w$ is in $P_k$ say $w=v_j$, then $j \le k-2$, and $v_jv_{j+1}\ldots v_kv_j$ is your desired cycle. If $w$ is not in $P_k$ then setting $w \doteq v_{k+1}$, we note that $P_{k+1} \doteq v_1\ldots v_kv_{k+1}$ is a path with $k+1$ vertices.
So let us now use this: Let $v_1$ be any vertex, and let $v_2$ be a neighbor or $v_1$, and in turn let $v_3$ be a neighbor of $v_2$ that is not $v_1$. This is a path $P_3=v_1v_2v_3$ of 3 vertices. From $P_3$ we use the above and find either a cycle in $G$ or a path $P_4$ on 4 vertices. From $P_4$ we use the above to find either a cycle in $G$ or a path $P_5$ in $G$ on 5 vertices. And so on and so forth. So from the above, we will have found either a cycle or a series of paths $P_4,P_5,P_6,\ldots$ where each $P_{k+1}$ is a path on $k+1$ vertices. However, at some point we will actually find a cycle from the above, instead of an infinite series of longer and longer paths. Indeed, letting $n$ be the number of vertices in $G$, there is no path in $G$ with more than $n$ vertices. So for some $k\le n$, we will find from the path $P_k$ on $k$ vertices, not a longer path on $k+1$ vertices, but instead a cycle. This is somewhat subtle, so make sure you see this yourself.
**You can also use the Handshaking Lemma and to conclude that $G$ has at least as many edges as cycles. And then from this note that a graph with at least as many edges as vertices has a cycle, but I was not sure if you had learned these results yet.
A: As I understand the basic case is a simple graph.

If the degree of each vertex of a simple graph $G$ is at least $2$,
then graph $G$ has a cycle.

Proof 1 (owned by @Mike). Suppose the graph $G$ has no cycles. Let $v_1\ldots v_k$ be an arbitrary path in $G$ of length $k$, in which all vertices are different. Since $\operatorname{deg}(v_k)\geq2$, there exists a vertex $v_{k+1}$ adjacent to $v_k$ and $v_{k+1}\neq v_i$ for all $i\leq k$. We have a path $v_1\ldots v_k v_{k+1}$ of length $k+1$. Since the graph has a finite number of vertices, we obtained a contradiction.
Proof 2. We can assume that graph $G$ is connected.
Suppose the contrary, a connected graph $G$ has no cycles. Then $G$ is a tree. But every tree contains a vertex of degree $1$. Contradiction.
