"If a vertex appears $k$ times in an eulerian circuit, then it must have degree $2k$" - Why?

Question

I need help understanding this part of this proof from Graphs and Digraphs (7th ed):

Theorem 3.1. A connected multigraph is eulerian if and only if every vertex has even degree.

The authors of this book consider a (multi)graph to be eulerian if it contains an eulerian circuit, i.e. a circuit that contains every edge of the graph.

Proof. One direction is clear. If a vertex appears $k$ times in an eulerian circuit, then it must have degree $2k$. Thus every vertex of an eulerian multigraph must have even degree.

The problems that I am having with this proof are (1) I don't know if I am interpreting this proof correctly and (2) it does not seem rigorous enough for me to use in an exam.

My intuition for this direction of this proof is that whenever you "pass through" a vertex in an eulerian circuit, you must be able to come out--so, for every edge incident to some vertex, there must exist another incident edge to that vertex. Again, however, I am doubting whether this is correct.

I have tried to formalize this by trying to think of an Eulerian circuit $C: v_1, v_2, \ldots, v_n$, where by definition we have $v_1 = v_n$ and that this contains every edge from some graph $G$. However, I don't see a way of deriving the fact that all vertices must be even from this.

Answer 1

Break the edges of the graph into half-edges and start the circuit in the middle of an edge. Each visit to a vertex removes two half-edges from further use - the incoming one and the outgoing one. So the number of usable half-edges at the vertex decreases by $2$. Its parity (even or odd) does not change. At the completion of the circuit, every half-edge has been used, so the number of usable half-edges at each vertex is now $0$, an even number. Thus all the parities of usable half-edges at any time must have been even. This includes before the first visit, when it was the degree of the vertext.