Proof
(related to Lemma: Splitting a Graph with Even Degree Vertices into Cycles)
 By hypothesis, \(G(V,E)\) is a simple graph in which each vertex has even degree.
 Obtain a fist cycle as follows:
 Denote \(G(V,E)\) by \(G_1(V,E_1)\).
 Start at any vertex \(v\) and traverse edges in an arbitrary manner, never repeating an edge. This is possible since the even degree of each vertex ensures that whenever we enter a vertex, we must be able to leave it via a different edge.
 Since \(V\) is finite, we must eventually reach a vertex \(v\) that we have visited before.
 The edges of the trail between the two occurrences of the vertex \(v\) must, therefore, form a cycle \(C_1\).
 Now remove from \(E_1\) the edges of \(C_1\), i.e. this leaves a graph \(G_2(V,E_2)\) with \(E_2:=E_1\setminus E(C_1)\).
 Claim: If \(G_2\) has any edges left, each vertex in \(G_2\) has still an even degree, since:
 We have removed exactly \(2\) edges from each vertex contained in \(C_1\): one via which we have visited it, and one via which we left it.
 Therefore, the even degree each vertex contained in \(C_1\) remains even in \(G_2\), since it was also even in \(G_1\).
 If \(G_2\) as any edges left, we can repeat the procedure above to find a cycle \(C_2\) in \(G_2\) with no edges in common with \(C_1\).
 This leaves a graph \(G_3(V,E_3)\) with \(E_3:=E_2\setminus E(C_2)\).
 Continue in this way until there are no edges left, at which stage the constructed cycles \(C_1, C_2, C_3,\ldots\) together include every edge of \(G\), and no two of which have any edges in common.
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References
Bibliography
 Aldous Joan M., Wilson Robin J.: "Graphs and Applications  An Introductory Approach", Springer, 2000