Let $G=(V,E)$ be a $2d$ -regular graph $d \ge1$, then $G$ has a 2-factor
If I got my definitions correct, a 2-factor is simply a subgraph $H=(V,F)$ of $G$ so that $d_H(v)=2$ for all $v \in V$.
Since $G$ is a $2d$ regular graph, there exists an eulerian cycle $C$ with edges $e_1 ...e_n$
I know that the alternating edges $e_1,e_3,e_5,,,$ form a k-factorization
How can I go from here ?
Rather than taking alternating edges, orient the edges of the Eulerian tour, so that every vertex has $d$ directed edges going into it and $d$ going out.
As always, the $n$-vertex directed graph corresponds to a $2n$-vertex bipartite graph, where both partite sets are copies of $V(G)$, and every directed edge $xy$ turns into an edge from the first copy of $x$ to the second copy of $y$. In this particular case, we get a $d$-regular bipartite graph.
A regular bipartite graph contains a perfect matching. If we take the edges of that perfect matching, they form a $2$-factor in the original graph.
(P.S. Actually, this only works on each connected component of $G$, but if we have a $2$-factor in each connected component, then together, they form a $2$-factor of the whole graph.)