Existence of d-regular subgraphs in a k-regular graph The claim is as follows:
Let's say we have a $k$-regular simple undirected graph $G$ on $n$ vertices. Then, does $G$ then always have a $d$-factor for all $d$ satisfying $1 \le d \lt k$ and $dn$ being even.

I think its true, since we can construct a


$k+1$-regular graph from $G$ by adding $\frac{n}{2}$ edges for even $n$ or 
$k+2$-regular graph from $G$ by adding $n$ edges for odd $n$


And the converse for the above.
We can construct a $k$-regular graph on $n$ vertices by


removing $\frac{n}{2}$ edges from a $k+1$-regular graph for even $n$. 
removing $n$ edges from a $k+2$-regular graph for odd $n$. 


Hence, using the converse argument, we can say that the original claim is true, since from a $k$-regular graph $G$ we can construct 


$1$-factor, $2$-factor, ... $k-1$-factor for even $n$ or
$2$-factor, $4$-factor, ... $k-2$-factor for odd $n$


Is my argument correct or is there any flaw in this argument?
Your conjecture is false. There are regular graphs with an even number of vertices yet without a 1-regular subgraph. See this question on Mathematics.

The complement of such a graph gives a counterexample to your claim that you can always add a perfect matching to increase the regularity (when the number of vertices is even).

In the bipartite case, however, your conjecture is true. Any $k$-regular bipartite graph can be decomposed into $k$ perfect matchings. This follows from Hall's theorem.