Partition of the edges of a $4$-regular graph

The edges of every $4$-regular non-directed graph can be divided into two sets, so that removing any vertex with its edges removes two edges from one set and two edges from the other.

Could someone please help me understand why this is the case?

Since the degree of each vertex is 4 (even), there exists an Euler circuit. Pick one and a starting-ending vertex, $S$. Label the first edge you traverse as you trace out the Euler circuit $a$, the second edge you traverse $b$, the third edge you traverse $a$, the fourth edge you traverse $b$, and continue labeling edges in this manner until you complete the circuit and get back to $S$.

By the Degree Sum Formula, $\Sigma d(v) = 2e$, so $e=\frac{\Sigma d(v)}{2} = \frac{4v}{2} = 2v$ (where $v$ is the number of vertices and $e$ is the number of edges). So there are an even number of edges and the last edge we traverse must be labelled $b$.

Now, pick a vertex that is not the starting-ending vertex $S$. Following the Euler circuit we must arrive at that vertex along an edge labelled $a$ and then immediately depart that vertex along an edge labelled $b$ (or vice-versa) and this must happen exactly twice at that vertex. As such, removing that vertex and the edges incident to it will remove two edges labelled $a$ and two edges labelled $b$.

Similarly, for the starting-ending vertex $S$ the Euler circuit will depart $S$ by traversing an edge labelled $a$, arrive back at $S$ and then immediately depart $S$ along edges labelled $a$ and $b$ (or vice-versa), and finally return back to $S$ along an edge labelled $b$. So again, removing $S$ and the edges incident to it will remove two edges labelled $a$ and two edges labelled $b$.