Worked solution: Euler's degree argument (1736)

Suppose a walk crosses every edge of the graph exactly once. Fix a vertex $v$ and look at every time the walk visits $v$ in the middle of the trip (that is, $v$ is neither the very first nor the very last vertex of that particular visit). Each such visit uses one edge to arrive at $v$ and a different edge to leave $v$ again, and because the walk never repeats an edge, no edge at $v$ is ever reused across different visits.

This pairs off the edges at $v$ into (arrival, departure) pairs, two at a time, for every visit that is not the overall start or overall finish of the walk. If $v$ is never the start or finish, all of its edges get paired this way, so $\deg(v)$ must be even. If $v$ is the start (and not also the finish), one departure edge is left with no matching arrival, so $\deg(v)$ is odd; symmetrically, if $v$ is only the finish, one arrival edge is left unmatched. If $v$ is both the start and the finish (a closed tour), the leftover departure and the leftover arrival pair with each other, and $\deg(v)$ is even again.

Because a single walk has exactly one starting vertex and exactly one finishing vertex — possibly the same one — this pairing argument shows directly that at most two vertices of the graph can end up with odd degree, and if the walk is closed, none do. This is the actual mechanism behind Euler's rule, not just a fact to take on faith.