Let $R$ be a unital ring, $a\in R$ and $M$ be a right $R$-module. Define the $$r(a)=\{x\in R:ax=0\},$$ the right annihilator of $a$. If $x\in r(a)$, show that $$Mx\subseteq \text{Ann}^{M}(a).$$ Where $\text{Ann}^{M}(a)=\{m\in M:ma=0\}$. (The confusion I get is that the only correct statement is $Ma\subseteq \text{Ann}^{M}(x)$), but below is the proof.
Attempt: Since $x\in r(a)$ implies that $ax=0$, then $0=a(xR)$ so that $xR\subseteq r(a)$. This implies that $Mx\subseteq \text{Ann}^{M}(a)$.
But I am not so sure of the working.
Counterexamples
1) Take $R=M_2(k)$, where $k$ is any unital commutative ring.
2) Take $M=k\times k$ with usual structure of $M_2(k)$-right module. Put $a=\left(\begin{array}{cc}1&0\\ 0&0 \end{array}\right)$ and $x=\left(\begin{array}{cc}0&0\\ 1&0 \end{array}\right)$. We have $ax=0$, that is, $x\in r(a)$. Moreover
$Mx=(k,0)$ and $(Mx)a\neq(0,0)$. Therefore $(Mx)\not\subset Ann^M(a)$.