I have devised a proof for this problem, which is based on Veblen's Theorem, which states that the set of edges of a finite graph can be written as a union of disjoint simple cycles if and only if every vertex has even degree.
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Let $G$ be the graph, with $n$ vertices. The degree condition is relaxed here, that is, the vertices can have odd or even degrees.
Now, a graph can be written as a union of edge-disjoint cycles if all the vertices have even degree.
By induction on $n$, we delete an edge from all vertices of odd degree, thus making the degree of the vertices even.
If any vertex of even degree becomes a vertex of odd degree as a result of deletion of edges, we further go on deleting edges until the whole reduced graph becomes a graph having vertices of only even degree.
Now, this graph
$G_1$ = $G$ - {$\cup e_{i}$}
has all vertices of even degree, and hence can be represented as a union of edge-disjoint cycles.
So, we can infer that every cycle in $G_1$ can also be represented as union of edge-disjoint cycles.
Now, we can permute the edges to be deleted and we get a new sub-graph with every vertex having even degree, with every permutation. The combination of edges to be deleted need to be chosen such that deleting them results in a sub-graph having every vertex with even degree.
Now, we add back the the deleted edges to each sub-graph to get back the original graph $G$.
Since, these edges do not contribute to any cycle in $G_1$, hence the cycles in $G_1$ are same as in $G$.
In this way, we get different cycles from different sub-graphs, obtained by deleting different combination of edges.
Now, we can say that, $G = \cup G_i$
Hence, the elements of cycle space in $G$ can be represented as union of edge-disjoint cycles in $G$.
Is this proof valid and logical??