For a fixed element of a ring R, show that the set is a subring of R. The question follows. For a fixed element of a ring R, show that the set $S=\{x\in \mathbb{R}|ax=0\}$ is a subring of R.
To show that S is a subring of R, it must meet the following conditions:
$(i) S \neq \emptyset$
$(ii) x \in S$ and $y \in S$ imply that $x-y$ and $xy$ are in S.
This is my proof:
$(i) $ Since S is a subring of R, then the additive identity is in S. Thus $a*0=0 \rightarrow 0=0$ and $0\in S$. Thus $S \neq \emptyset.$
$(ii)$ Let $x,y \in S$, then $ax=0$ and $ay=0$. Thus $a(x-y)=0 \rightarrow ax-ay=0 \rightarrow 0-0=0.$ Thus $x-y\in S.$ Which also following with $xy$.
Let $x,y \in S$, then $ax=0$ and $ay=0$, then $a(xy)=0 \rightarrow (ax)y=0 \rightarrow 0*y=0 \rightarrow 0=0$, thus $xy \in S.$
Is it right? any modifications that I can do to make it better?

You have the right ideas, you just need to write them down better. For instance, you want to prove that $S$ is a subring, so you cannot start with “since $S$ is a subring of $R$”. Also the argument in (ii) is backwards.
Avoid those pesky arrows! ;-) It seems that you start from $a(xy)=0$ (which instead is what you need to prove) and deduce that $0=0$. But $0=0$ can be deduced from everything, even from a false assumption!
Revised version
(i) Since $a0=0$, we see that $0\in S$, so $S\ne\emptyset$.
(ii) Suppose $x,y\in S$. Then $ax=0$ and $ay=0$ by definition; so
$$
a(x-y)=ax-ay=0-0=0
$$
and therefore $x-y\in S$. Similarly,
$$
a(xy)=(ax)y=0y=0
$$
which proves $xy\in S$.
