Please note that I am not looking for a solution to the question in the title as it has been asked before.
I was trying to prove this famous result to a friend:
If $I$ is an ideal of a ring $R$ and $P_1, P_2$ are prime ideals of $R$ such that $I\subseteq P_1\cup P_2$, then $I\subseteq P_1$ or $I\subseteq P_2$.
The trouble is, I forgot the proof, and came up with a bizarre result that I would like feedback on.
My proof attempt:
Suppose $I\subseteq P_1\cup P_2$. Suppose towards a contradiction that there exist $a\in I\cap (P_1\setminus P_2)$ and $b\in I\cap (P_2\setminus P_1)$. Then $a+b\in I$ since $I$ is an ideal. If $a+b\in P_1$, then $a+b-a=b\in P_1$, a contradiction. If $a+b\in P_2$, then $a+b-b=a\in P_2$, a contradiction. Thus it is not the case that there exist such $a, b$, and hence $I$ is completely contained in $P_1$ or $P_2$.
There are a few problems with this. Namely, I did not use the fact that $P_1$ and $P_2$ are prime, and I think this is quite obviously not true for ideals in general. But after investigating the proof, I don't quite see a problem with it.
Naturally, this brings us the question: what is wrong with this proof? Is the result actually true for two ideals in general? Why do we need primeness in any case?