Handshake Lemma

Theorem
The number of vertices with odd degree in a graph must be even.

Direct Proof
Let $$G = (V,E)$$ be an unweighted, undirected graph. Consider the sum of the degrees of its vertices: $$K = \sum_{v \in V} deg_G(v)$$.

Each edge $$e \in E$$ will be counted exactly twice by this sum, once for each vertex to which it is incident, so this sum must be equal to the twice the number of edges in $$G$$. That is, $$K = 2|E|$$, an even number.

Subtracting from $$K$$ the degrees of all vertices of even degree, we are left with the sum of all degrees of vertices in $$V$$ with odd degree.

That is, $$\left( \sum_{v \in V} deg_G(v) \right) - \left( \sum_{ \{v \in V : 2 \mid deg_G(v) \} } deg_G(v) \right) = \left( \sum_{\{v \in V : 2 \nmid deg_G(v) \} } deg_G(v) \right)$$.

We are left with the sum of degrees of vertices of odd degree, which still must be an even number, as it is equal to the difference of two even numbers. (We know that the first sum is even, and the sum of degrees of vertices with even degree will clearly be even as well.)

Since this is a sum of exclusively odd terms, there must be an even number of such terms for the sum on the right side to be even, so we have an even number of vertices of odd degree, as required.