I want to check my proof:
Given: Let $R$ be a ring with $1$ and $H$ an additive subgroup of $R$. We define $R_0=\{x\in R : \forall h \in H \text{ we have } xh \in H\} \subset R$.
To prove: $R\neq \{0\} \Longrightarrow R_0 \neq \{0\}$.
(I showed that $R_0$ is a subring of $R$ separately, but I don't see how this would be relevant to what is asked here.)
My attempt: I wanted to show that $R_0=\{0\} \Longrightarrow R=\{0\}$.
Starting with $R_0=\{0\}$, we have that the only element $x$ in $R$ such that $xh\in H$ for all $h\in H$ is $x=0$. This implies that $0=1$, since $1\in R$ and $1h=h\in H$ for all $h\in H$.
I don't think this is enough to conclude that $R=\{0\}$.
So let's take some $a\in R$ such that $a\neq 0$. Since $H$ is an additive subgroup of $R$, we have that $ah \in H$ for all $h\in H$, since $R$ is closed under the ring operations. So contradiction: now $R_0=\{0,a\}$. So there is no such $a\in R$. So $R=\{0\}$. This finishes the proof.