Proving that $\sum \deg(v) = 2m$ for any Graph $G$ Here is My proof, please correct me if wrong, I try to be formal.
Proof by Induction:
Let $\sum \deg(v)=2m$ assumption... when #of nodes is $n=0$.
so here the equation is  $\sum \deg(v)=2(0)=0$ [true for 0] .... (1)
now if we add new node $w$, and connect it to node v then we expect that only the degree of $v$ and $w$ will increased by $1+1 =2$ , and 
so doing that like the following:$\sum \deg(v)=2 + 2m = 2(m+1)$,
then our assumption is true because when we add 1 edge, total  number of edges increased by $2(m+1)$.
Is it correct and formal proof ?
 A: As written this only proves the proposition for connected graphs without cycles.  In your inductive step you add a new vertex and edge to an existing graph.  This restricts the overall types of graphs.  However, it is mostly correct, here is how I would modify it.
First, instead of inducting over the number of vertices in a graph $n$, use the number of edges $m$.  Now your inductive step can look like this: Suppose we have a graph $G$ with $m$ edges.  Remove one edge at random.  The resulting graph $G'$ has $m-1$ edges, and by induction, we know the proposition holds for that graph.  Therefore, $\sum \deg(v) = 2(m-1)$ for the subgraph $G'$.  $G$ is just the graph $G'$ with one more edge.  Since this edge is connected to two vertices in $G$, the degree of each of those vertices has now increased by one.  So $\sum \deg(v) = 2(m-1) + 2 = 2m$ for the graph $G$.
That would yield a full formal proof.
Update
For the base case you will need to prove the proposition for all graphs with $0$ edges.  Namely the graphs that consist of $0$ or more vertices and no edges.
