Proof of the degrees along an Eulerian trail

Proof

We count, for each vertex, how often the trail enters and leaves it.

1. Write $W = (w_0, w_1, \dots, w_k)$ with $w_0 = u$ and $w_k = v$. It is a trail, so its edges $w_{t-1} w_t$ for $1 \le t \le k$ are pairwise distinct.
2. Since $W$ is an Eulerian trail, every edge of $G$ is one of these $k$ edges, and each appears exactly once.
3. If $k = 0$, then $G$ has no edges and $u = w_0 = v$. Every vertex has degree $0$, which is even. So the statement holds. From now on let $k \ge 1$.
4. Fix a vertex $x$. For a condition $\varphi$, let $[\varphi]$ be $1$ if $\varphi$ holds and $0$ otherwise.
5. The degree of $x$ is the number of edges that contain $x$. By steps 1 and 2, $\deg(x)$ is the number of $t \in \set{1, \dots, k}$ with $x \in \set{w_{t-1}, w_t}$.
6. In a graph, $w_{t-1} \ne w_t$, because adjacent vertices are distinct. So $[x \in \set{w_{t-1}, w_t}] = [w_{t-1} = x] + [w_t = x]$.
7. Summing over $t$ gives $\deg(x) = \sum_{t=0}^{k-1} [w_t = x] + \sum_{t=1}^{k} [w_t = x]$.
8. Hence $\deg(x) = 2 \sum_{t=1}^{k-1} [w_t = x] + [w_0 = x] + [w_k = x]$.
9. So $\deg(x) \equiv [x = u] + [x = v] \pmod 2$.
10. If $u = v$, the right side of step 9 is $2[x = u]$, which is even. So every vertex has even degree.
11. If $u \ne v$, the right side of step 9 is $1$ for $x = u$ and for $x = v$, and $0$ for every other $x$. So $u$ and $v$ have odd degree, and every other vertex has even degree.