Let $A$ be a semisimple algebra and $e$ be an idempotent in $A$. Then $eAe=\{eae:a\in A\}$ is a subalgebra of $A$ with $e$ as the identity. We want to prove that $eAe$ is also semisimple. That is, if $\sigma_{eAe}(ax)=\{0\} \forall x\in eAe$, we must show that $a=0$.
We also know that $\sigma_{A}(x)=\sigma_{eAe}(x)\cup\{0\}$. Hence if $\sigma_{eAe}(ax)=\{0\} \forall x\in eAe$, we have $\sigma_{A}(ax)=\{0\} \forall x\in eAe$ or equivalently, $\sigma_{A}(aexe)=\{0\} \forall x\in A$.
How can we conclude from this that $a=0$ using the fact that $A$ is semisimple?
I'm not familiar with the characterization of semisimple you use. But, assuming that we use "$A$ semisimple" to mean that $\sigma_A(by)=\{0\}$ for all $y$ implies $b=0$:
If $a\in eAe$ and $\sigma_{eAe}(ax)=\{0\}$ for all $x\in eAe$, then as you say $\sigma_A(aexe)=\{0\} $ for all $x\in A$. But then $\sigma_A(eaex)=\{0\} $ (because $\sigma(xy)\cup\{0\}=\sigma(yx)\cup\{0\} $).
As $A$ is semisimple, this implies that $eae=0$. Then $a=eae=0$.