We want to proof $2|E| = \sum \limits_{v \in V} deg(v)$ for a simple graph (no loops). For our proof we assume $n$ to be the number of edges in a simple graph $G(E, V)$. We proceed our proof by induction.
Base case P(0), no edges exist, so all nodes in $G$ have degree 0. Therefore we find that $2n = 2 * 0 = \sum deg(v) = 0$
Inductive step, assuming P(n) is true, we need to show that P(n + 1) is also true, that is:
$2(n + 1) = \sum \limits_{v \in V} deg(v)$
In a graph $G$ with number of edges $n + 1$. If we remove one edge at random $G$, we get a subgraph $G'(E',V')$ for which we can assume P(n):
$2n = \sum \limits_{v \in V'} deg(v)$
$G$ is equal to the subgraph $G'$ plus one edge. As every edge contributes $2$ to the total number of degrees (as every edge connects two vertices) we can say for $G$:
$2n + 2 = 2(n + 1) = \sum \limits_{v \in V'} deg(v)$
Which proofs P(n + 1).
Does the above proof make sense? I had a look at some other questions, but couldn't find a fully written proof by induction for the sum of all degrees in a graph.
Best Answer
Whenever an edge is introduced in a graph; It will connect two nodes (vertices). So degree of both those nodes will increase by $1$.
Thus Sum of degrees will increase by $2$.
So we can say that every addition of edge increases sum of degrees by $2$.
So if there are $E$ such edges: sum of degrees $= 2 + 2 + 2 \ldots$ ($E$ times) = $2E$.