Write such primes by picking up one variable at a time from the products $x_1x_2$, $x_3x_4$, $x_5x_6$, and so on. (Every prime ideal containing $(x_1x_2,x_3x_4,x_5x_6,\dots)$ must contain one variable from each product.) For example, one of such minimal primes is $P=(x_1,x_4,x_5,\dots)$. It is clear (I hope) that in this way you will find infinitely many minimal primes.
1). Having identity does not factor into the proof. Using noncommutativity would require you to restate everything in terms of right or left ideals.
2) Is an ideal generated by an infinite set has in fact infinitely many generators, or does this just mean that such an ideal is simply generated by infinitely many elements, but not necessarily infinitely many generators? The second thing is more correct. Really you should probably be thinking in terms of "finitely generated" and "not finitely generated" (which is a little different from "infinitely generated".)
3) The proof of the on direction is not correct as we have been discussing in the comments. To get you started for the $\impliedby$ direction:
Assume all ideals are finitely generated and let $I_1\subseteq I_2\subseteq\ldots$ be an ascending chain. Let $I=\bigcup_{i=1}^\infty I_i$ be the union of these ideals (and prove it is an ideal, if you haven't already.
Now $I$ is finitely generated by $\{a_1, a_2,\ldots a_n\}$. The important step in logic takes place right here:
prove that there must be a $j\in \mathbb N$ such that $\{a_1, a_2,\ldots a_n\}\subseteq I_j$. Conclude that $I=I_j$, and that $I_k=I$ for all $k\geq j$.
The idea for $\implies$ is the right one, but it has a technical flaw: you do not establish that the chain is strictly increasing. You can do this inductively. (The other posts mentions a fault with assuming $I$ is countably generated; however, I only see that as a disagreement in notation, and what is written doesn't really say that. You can well-order whatever generating set and write the initial segment of countably many generators like that, IMO.)
Best Answer
Hint: Consider the ideal $I=\langle x_1,x_2,x_3... \rangle$.
Following what anon said. You can apply the same trick one below to prove what annon is asking you.
Alternate: Let us prove that $R'=R[x_1,...]$ is not noetherian. Take the chain $(x_1)\subset (x_1,x_2)\subset\cdots$. If this stabilizes, then for some $n$ we have $(x_1,...,x_n)=(x_1,..,x_{n+1})$. Then $x_{n+1}=h_1x_1+\cdots+h_nx_n=f$. Evaluate $f$ at the point that is $0$ in every coordinate except $n+1$, here say it is $1$. Then you get $1=0$. A contradiction.
Remember that a different characterization of Noetherian is that every ideal is finitely generated. Then by above since it is not noetherian, we have that there must be an ideal that is not finitely generated.