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Assume a proper ideal $I_{1}$ of $R$ is not contained in any maximal ideal of $R$. Then $I_{1}$ is not maximal since $I_{1}$ contains itself. Then there is an ideal $I_{2}$ such that $I_{1}\subset I_{2}\subset R$. By assumption, $I_{2}$ is not maximal, so there exists an ideal $I_{3}$ such that $I_{1}\subset I_{2}\subset I_{3}\subset R$. By assumption, $I_{3}$ is not maximal. We continue the process of choosing ideals so that we obtain a strictly ascending infinite chain $I_{1}\subset I_{2}\subset \dotsc$ of ideals, which is a contradiction since $R$ is PID, so every strictly ascending chain of ideals of $R$ should be of a finite length.

As you can see here, I did not use the assumption that $R$ is a domain. $R$ being an arbitrary principal ideal ring should suffice to imply a contradiction. It's either I missed the part where I used the "domain", or there is a flaw in my proof, or the assumption that $R$ is a domain is not necessary at all. Can someone enlighten me?

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Your proof is correct. Some possibly-enlightening remarks:

  1. You have essentially just used that every ascending chain of ideals in a PID must stabilise.
  2. A ring in which every ascending chain stabilises is called Noetherian.
  3. An equivalent condition for being Noetherian is: Every ideal is finitely generated.
  4. In particular, any principal-ideal-ring is Noetherian and thus, every ascending chain stabilises. So you are correct in noting that that is all that you needed.

Assuming the axiom of choice, your statement is true for any non-zero commutative ring with unity.

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