If $R$ is Noetherian and $R/I$ is finite, then $R/I^2$ is finite. Why is being Noetherian necessary? What goes wrong for non-Noetherian R in this proof?
Suppose $a_1, \ldots, a_n$ are the representatives of $R/I$.  Then $(a_1+I)\cup\ldots\cup( a_n+I)=R$, so $(a_1+I^2)\cup\ldots\cup( a_n+I^2)=I$.  Then $$(a_1+a_1+I^2)\cup\cdots\cup( a_1+a_n+I^2)\cup( a_2+a_1+I^2)\cup\cdots\cup( a_2+a_n+I^2)\cup\ldots\cup( a_n+a_1+I^2)\cup\cdots\cup( a_n+a_n+I^2)=\\=a_1+(a_1+I^2)\cup\ldots\cup( a_n+I^2))\cup( a_2+(a_1+I^2)\cup\ldots\cup( a_n+I^2))\cup\ldots\cup( a_n+(a_1+I^2)\cup\ldots\cup( a_n+I^2)=\\=(a_1+I)\cup( a_2+I)\cup\ldots\cup( a_n+I)=R$$
Update: Thank you to both of you, who pointed out the reason my original proof didn't work.  I think now I will take the time to think about what the correct proof would look like.
 A: To see why the Noetherian hypothesis is necessary, let $k$ be a finite field and consider $R=k[x_1,x_2, \dots]$ (the polynomial ring in countably infinite variables) and $I=(x_1,x_2, \dots)$. Then we have $R/I \cong k$, so $R/I$ is finite. 
But on the other hand $I^2=(x_ix_j)_{i,j \in \mathbb N}$, so that $I^2$ is generated by all the homogenous polynomials of degree $2$. Now for $i \neq j$, $x_i-x_j$ is homogenous of degree $1$. If we suppose that $x_i-x_j \in I^2$, then $x_i-x_j$ could be written as a finite linear combination $x_i-x_j=\sum_k P_kH_k$, where each $H_k$ is homogenous of degree $2$, if we write each $P_k$ as a sum of homogenous polynomials and use the fact that the product of homogenous polynomials is again homogenous and that for any fixed degree, the space of homogenous polynomials is closed under addition, we see that this is impossible.
This means that for $i \neq j$ we have that $x_i-x_j \notin I^2$, so the elements $x_i+I^2$ and $x_j+I^2$ are distinct in $R/I^2$, thus $R/I^2$ has infinitely elements.
For the proof in the Noetherian case, I will give you a few hints (try to prove these first, if you are not familiar): 


*

*$(R/I^2)/(I/I^2) \cong R/I$

*If we have that $M/N$ and $N$ are finite, then $M$ is finite.

*If $M$ is a finitely generated $R$-module, then $M/IM$ is a finitely generated $R/I$-module

*A finitely generated module over a finite ring is finite as a set.

A: You only need to use that $I$ is finitely generated ( as an $R$ module). So $I/I^2$ is also finitely generated $R$ module, hence a finitely generated $R/I$ module. However, $R/I$ is finite, so we conclude $I/I^2$ is finite. So now $R/I$, $I/I^2$ finite implies $R/I^2$ finite. 
Note: If $I$ is finitely generated, $I^k$ will also be finitely generated for all $k\ge 1$. An easy induction now shows that all the rings $R/I^k$ are finite. 
