Quotients and localizations of an Artinian ring are Artinian

Let $I\subseteq R$ be an ideal. By Correspondence theorem: ideals of $R\mathrm{/}I$ correspond to ideals of $R$ containing $I$, descending chains of ideals in $R\mathrm{/}I$ correspond exactly to descending chains of ideals of $R$ containing $I$. Since those stabilize in the Artinian ring $R$, the quotient $R\mathrm{/}I$ is Artinian.

If $R=0$, then every localization is again $0$ and there is nothing to prove. Otherwise An Artinian ring is canonically the finite product of its localizations at its maximal ideals gives maximal ideals ${\mathfrak{m}}_1,\dots ,{\mathfrak{m}}_r$ such that $R\cong {\prod }_{i=1}^r{R}_{{\mathfrak{m}}_i}$. Localization of a finite product acts factorwise, so it is enough to localize an Artinian local ring $(A,\mathfrak{m})$. If $S\cap \mathfrak{m}\mathrm{\ne }\mathrm{\varnothing }$, choose $s\in S\cap \mathfrak{m}$. By An Artinian local ring has nilpotent maximal ideal, and its finite modules have finite length, $s^n=0$ for some $n$, so $(s\mathrm{/}1{)}^n=0$ in $S^{-1}A$ while $s\mathrm{/}1$ is a unit; therefore $1=0$ and $S^{-1}A$ is the zero ring. If $S\cap \mathfrak{m}=\mathrm{\varnothing }$, then every element of $S$ is a unit of $A$: otherwise $(s)$ would be a proper ideal and therefore lie in the unique maximal ideal $\mathfrak{m}$, contradicting $s\mathrm{\notin }\mathfrak{m}$. So the localization map $A\to S^{-1}A$ is an isomorphism. Hence a localization of $R$ is exactly the product of those local factors whose maximal ideals avoid $S$, with the other factors collapsing to zero.

A finite product of Artinian rings is Artinian, because a descending chain of ideals in the product is coordinatewise a descending chain in each factor and therefore stabilizes once every coordinate chain does. Step 1.1 handles quotients, and step 2.1 handles localizations.