Let $R$ be a non-commutative semiprime Ring with unity and $(0) \neq S$ be a two-sided Ideal of $R$ wich is a minimal Annihilator-ideal ( $S = l(X)$ for some left-ideal $X$ of $R$). Then $S$ is a subring of $R$ (without unity). Let $T \neq (0)$ be a left-ideal of $S$. Then $ST$ is a left-ideal of $R$. I want to show that $ST \neq (0)$.
We have $TT \subseteq ST = (0)$, but since $R$ is semiprime and $ T \neq (0)$, it is $(0) \neq T \subseteq RT $ i.e. $(0) \neq RTRT$. But I don't know really how to prove the claim.
In a semiprime ring, the left and right annihilators of an ideal coincide.
Then from $ST=\{0\}$ you know also $TS=\{0\}$, and hence $(RT)S=\{0\}$.
By (something very similar to) your observation, this would mean $(RT)^2\subseteq (RT)S=\{0\}$, so that $RT=\{0\}$ by semiprimeness. It follows that $T=\{0\}$, contradicting the hypothesis that $T$ isn't zero.
So, it can only be the case that $ST\neq\{0\}$.