Let $R$ be an artinian ring. Then
$R$ is semilocal.
$(\operatorname{rad}(R))^n=0$ for some $n$.
Proof: Let $I=\text{rad}(R)$. By \hyperref[3.15]{3.15}, $\exists n$ so that we have $\operatorname{ann}_{\overline{R}} \overline{I}=\overline{0}$, where we denote $\overline{I}\subset \overline{R} = R / (0:I^n)$. We claim $\overline{R}=0$. ($\Rightarrow 0:I^n=R \Rightarrow I^n=0$).
Suppose $\overline{R}\neq 0$, then since $\overline{R}$ is artinian, $\exists N$ minimal with the property $N\neq 0$. Now $N$ is simple. Thus $N$ is finite, and $\overline{I}N=0$ or $\overline{I}N=N.$ The first equation is ruled out by \hyperref[3.15]{3.15}, the second one by \hyperref[2.2]{2.2} Nakayama's Lemma, ($I=\text{rad}(R)$), $N$ finite $R$-module.
I am going over my notes from last semester and I am trying to make the distinction of the proof of 1st bullet point and the 2nd (if there is one). Can someone help me with that? I am providing below the results that I am referring to.
3.15: Let $R$ be a ring that is artinian or noetherian. $I$ an $R$-ideal. Then $0:I^n=0:I^{n+1}$ for some $n$. For such $n$, write
$ \overline{R} = \left. R \middle/ (0:I^n) \right.$ and $\overline{I}=IR$.
Then $\operatorname{ann}_{\overline{R}} \hspace{0.1cm} \overline{I}=0$.
2.2: (Nakayama's Lemma) $M$ a finite $R$-module, $I$ an $R$-ideal. If $M=IM$, then $aM=0$ for some $a\in 1+I$. In particular, if $M=IM$ and $I\subset \operatorname{Rad}(I)$, then $M=0$ (since $a\in R$).
Everything you have written seems to be geared to proving the second point. I see nothing about the first point.
The first point is trivial though, so maybe you didn’t write it down? $R/J(R)$ has Jacobson radical zero, and is artinian since $R$ is. Therefore it’s semisimple, so $R$ is semilocal.