Let $P$ be prime and contain $IJ$, the product ideal. Then $I \subset P$ or $J \subset P$ So I already witness the solution. It is this: Assume $I \not\subset P$, then there is $i \not\in P$. Then the product $ij \in IJ \subset P$, but since $P$ is prime, $i \in P$ or $j \in P$, so $J \subset P$.
here is what I don't get. Doesn't this only show a finite truncation belongs to $P$? Don't we have to show that for any finite sum $i_1j_1 + \dots + i_nj_n \in P$, $i_n \in P$ for every $n$ (assuming $|J| = |I| = n$ for ease of argument)

The proof as written is phrased a little awkwardly and missing important details.  Here is a correct proof (with everything spelled out): 
Suppose $IJ \subset P$ and $I \subsetneq P$. We wish to show that for all $j \in J$, we have $j \in P$. Fix $j \in J$ $i \in I \setminus P$ and note that $ij \in IJ$. Since $IJ \subset P$, we have that $ij \in P$ but $P$ is prime, so we must have that either $i \in P$ or $j \in P$. Since $i \notin P$ (remember, $i \in I \setminus P$), we conclude that $j \in J$. This shows that $J \subset P$.
As far as the finite sums go, they belong to $IJ$ and thus to $P$ by assumption.
