Property of a connected graph with $\text{deg(all nodes)}=\text{even}$

Question

In my lecture nodes it is proposed that for a connected graph $G$ with the degree of all nodes being even, there exist two paths between any two nodes $x,y\in G$ with no common edges. From the very definition of a connected graph, there must exist one path between $x$ and $y$; the reason why there must exist two paths is then because we can simply "walk in the opposite direction" to get a new path? Also, how can it be shown that the two paths considered contain no common edges?

Answer 1

Since all degrees are even and the graph is connected, it has an Eulerian circuit. Starting at $x$ and walking clockwise around the Eulerian circuit gives one trail from $x$ to $y$. If this trail has repeated vertices, they can be "short-circuited" until only a path remains.

Starting at $x$ and walking counter-clockwise gives a different path from $x$ to $y$, with no edges in common with the first, because all edges in the circuit are distinct.

Answer 2

Notice that for every node $x$ and $y$, $x,y$ are nodes in a cycle. Then, we can easily "split" that cycle to get two $x,y$-paths that share no edge in common.