Let $G$ be a graph, If $G$ does not have vertices of odd degree, then $G$ cannot be a tree and hence contains a cycle. Now we prove by induction of $|E(G)|$ than $E(G)$ can decomposed as disjoint union of circles.
Base case: $|E(G)|=2$ is trivial to check.
Now, Take a connected component $Co_1$ of $G$, $Co_1$ has a circle $C_1$ by previous reasoning. Now, $G-|E(C_1)|$'s edges can be decomposed as disjoint union of circles $C_2,...C_n$ by induction hypothesis, Well those circles together with $C_1$ form a disjoint union of circles for $E(G)$. Done.
Here is the proof from some notes.