Let $a$ belong to a ring R. Let $S=\{x \in R | ax=0\}$. Show that S is a subring of R. Let $a$ belong to a ring R. Let $S=\{x \in R | ax=0\}$. Show that S is a subring of R.
My approach is such:
Let $c,d \in S$ so $(c-d)x=cx-dx=0-0=0 and (cd)x=(cx)d=(0)d=0$. Therefore by the subring test, S is a subring of R. Q.E.D
Is this correct or not so much?

You didn’t look carefully enough at the definition of $S$: $a$ is a fixed element of $R$, and to show that some $r\in R$ belongs to $S$, you must show that $ar=0$.
Suppose that $x,y\in S$. Then $a(x-y)=ax-ay=0-0=0$, so $x-y\in S$. Can you finish up by showing correctly that if $x,y\in S$, then $xy\in S$? (And remember, you can’t assume that $R$ is commutative, as you did in the step $(cd)x=c(xd)$ in your first attempt.)
