I am currently working on showing this:
Let $R$ be a semisimple ring, and $e\in R$ be idempotent, then if $eRe$ is a division ring, $Re$ is a simple ideal.
I am unsure if my working so far is correct. What I have done so far is the following.
Let $I\subseteq Re$ be a left ideal of $R$. Since $R$ is semisimple, $I$ is isomorphic to a simple $R$-module. If $x\in I$, then $x=re$ for some $r\in R$, and so $ex=ere\in eRe$. If $ex\neq0$, then $ex$ has an inverse in $eRe$, as $eRe$ is a division ring, and so $(ere)(er'e)=e$ for some $r'\in R$. Thus, $r\cdot r'=r'\cdot r=1_{R}$. This then gives us that $e=r'\cdot x\in Rx\subseteq I$. This then means that $Re\subseteq I$, and so $I=Re$, which means that $Re$ is simple.
I don't know how to consider the case where $ex=0$, whether this is ruled out or not by the fact that $R$ is semisimple. I am also unsure if my reasoning for $I=Re$ follows correctly, so any hints/advice would be helpful!
Also, I am aware that this post is similar, but cannot seem to follow it fully: eRe is a division ring implies eR is a minimal ideal
Here's one way that'll work.
It's worth knowing that $(eRe)^{op}\cong End(Re)$ via the map $exe\mapsto (re\mapsto reexe)$ Unfortunately we need the opposite mapping because of the choice of left $R$ modules. But fortunately $eRe$ is a division ring iff $(eRe)^{op}$ is, so it makes little difference to us.
Now, using the last paragraph, one can note that if $Re$ is not a simple left ideal, then $Re=A\oplus B$ for two nonzero left ideals. This decomposition corresponds to a nontrivial idempotent in $End(Re)\cong (eRe)^{op}$. But being a division ring, it should not have any nontrivial idempotents. Hence $Re$ is a simple left ideal.
Here is the earlier attempt which I wish I had a very elementary completion to. It relies on showing that if $re$ is nonzero, $ere$ is nonzero. I'd like to put it on the shelf for now because the method above seems preferable in this case.
There is a well-known characterization of simple modules that might help you: A module $M$ is simple if for any two nonzero elements $x,y\in M$, there exists an $r\in R$ such that $rx=y$.
Let $re$ and $se$ be any nonzero elements of $Re$. Let's also keep in mind that $e$ is the identity of $eRe$.
In $eRe$, find $ere$'s inverse (if $ere\neq 0$... it should be?), call it $ete$. Then $se(ete)re = se(etee)re = se(ete)(ere)=see=se$. This shows $Re$ satisfies the criterion mentioned above.