I don't quite understand how this proof from Eisenbud goes. If induction on $n$ is used, one should assume that the result holds for $n=k-1$ ($k>1$) and prove it for $n=k$. But the actual induction hypothesis assumes that the lemma fails for $n=k-1$. If the proof is by contradiction and by induction, shouldn't one assume that the contrapositive is true for $n=k-1$ and assume the converse for $n=k$ and arrive at a contradiction?
If what I said above doesn't make sense, I can restate the question as "please clarify the proof of Lemma 3.3." (I don't understand what's happening in general...)
Lemma 3.3. Suppose $I_1,\dots,I_n,J$ are ideals of a ring $R$, and suppose that $J\subset \cup_j I_j$. If $R$ contains an infinite field or if at most two of the $I_j$ are not prime, then $J$ is contained in one of the $I_j$.
Proof. (I omit the case when $R$ contains an infinite field.) Induction on $n$. The case $n=1$ is trivial. By induction, we may suppose that $J$ is not contained in any smaller union of the $I_j$, so we can find elements $x_i\in J,x_i$ not in $\cup_{j\ne i} I_j$. Supposing that $J\subset \cup_j I_j$, we must have $x_i\in I_i$. If $n=2$, then $x_1+x_2$ is in neither $I_1$ nor $I_2$, contradicting the supposition. If on the other hand $n>2$, then we may assume that $I_1$ is prime, and $x_1+x_2x_3\dots$ is not in any of the $I_j$, again a contradiction.