Suppose $R$ is a ring and $I$ is a prime ideal of $R$. Show that if $J,K$ are left ideals of $R$, then $JK \subset I \implies J \subset I$ or $K \subset I$.
I am given a hint to show that the set $A=\{a \in R \mid aK \subset I\}$ is an ideal containing $J$. I did this and will spare the details unless requested.
Now we suppose $j \in J-I$. Then $JK \subset I$ implies $jK \subset I$. So for all $k \in K$ we have $jk=i$ for some $i \in I$. Since $I$ is prime, $K \subset I$.
Now suppose $k' \in K-I$. Then $k'J \subset J \subset A$ so for all $k \in K$ and $j \in J$, $k'jk \in I$. In particular $(k'j)k' = i$ for some $i \in I$. Then by primeness again we see that $k'j \in I$ and finally $j \in I$ for all $j \in J$.
This looks convincing (if not rather inelegant) to me but why did I go though the trouble of showing $A$ was an ideal?
As it looks like you began to suspect at the end of your comments, you are not using the definition of 'prime' appropriate to the context. For noncommutative rings, here are versions of the definition of prime.
In the third paragraph, you assume the commutative definition of a prime ideal, which is too strong, and so the conclusion does not follow (for noncommutative rings.)
From the hint given, I believe you want to use the basic definition of a noncommutative prime ideal, which is that $I\neq R$ and that the product of two ideals being contained in $I$ implies that one of the ideal is contained in $I$. That means the task here is to prove one of the equivalences in the link above, that you can replace two-sided ideals with one-sided ideals.
So, you proved that $A=\{a\mid aK\subseteq I\}$ is an ideal containing $J$. We also have $AK\subseteq I$. Then by the same token $B=\{b\mid Ab\subseteq I\}$ is an ideal containing $K$ such that $AB\subseteq I$. Now apply the basic definition of a noncommutative prime ideal, and you are basically done.