Your proof would be absolutely fine if we can prove that the construction of a simple edge disjoint cycle is always possible. Take any vertex u with a positive degree and start a tour on adjacent vertices until you are unable to explore further. Keep deleting the edges you have traveled. Assume you end at vertex v. What we have is a trail (not path):
$$u,u_{1},u_{2}........v$$
The claim is, v has to be u.
Assume u and v are different. For every appearance of vertex v except the last one in the above list, there need to be two edges incident on v, one incoming and other outgoing. But for the last appearance, we need only one edge to enter the vertex v. Hence odd number of edges incident on v are used in the above trail. This leads to the fact there is at least one more edge on v yet to be explored. So the vertex v can not be the end of traversal. This proves u=v.
Now we have the circuit, which can be broken into disjoint edge cycles. How? For every two occurrence of a vertex y in the circuit, remove the circuit y-y from the parent circuit. Keep doing this recursively to get cycles.
Example: circuit $u,u_{1},u_{2},u_{3},u_{2},u$ be broken into $u,u_{1},u_{2},u$ and
$u_{2},u_{3},u_{2}$
Since we have deleted the edges explored, the cycles have disjoint edges. We recursively keep finding circuits in the main graph, deleting the edges until we have explored all the edges. Which completes your proof.
An easy proof- let $G_{1}$,$G_{2}$,$G_{3}$... be connected components of the graph. Since the degree of each vertex is even there are Eulerian circuit for each component. Then the trails can be broken into disjoint cycles for each component.
Best Answer
If by even graph you mean all vertices have even degrees then you do as follows: start at any vertex and keep on walking, until you hit a vertex you already visited. That means you have a cycle. Remove the edges of that cycle from the graph. The remaining graph is still even. Proceed by induction.
The only thing you must notice is that "keep on walking" is well-defined, i.e. you won't get stuck, but that follows from the fact that the degree of each vertex is at least $2$. So if you entered a vertex, you can always find (some other) exit.