Let $R$ be a unital ring and let $X=\{I_{\alpha} : \alpha \in A\}$ be a totally ordered (by subset) set of proper ideals of $R,$ indexed by some set $A.$ Prove that $I=\bigcup_{\alpha\in A} I_{\alpha}$ is a proper ideal of $R.$
By the definition of a proper ideal, I need to show that there is some element in $R$ that is not in $I$ and that $I$ is an ideal of $R.$ But I'm not sure how to arrive at this answer using just the information given in the question.
Edit: my previous answer was complete gibberish. I tried to use Zorn's lemma because the question looked similar to one that I solved using the lemma, but it turns out that that approach is completely flawed, as pointed as in the comments and the answer.
I don't see at all how any of the things you wrote proves that $I$ is a proper ideal. Here is how you usually prove such things:
We first show that $I$ is a proper subset of $R$.
Assume to the contrary that $R=\bigcup_a I_a$. In particular, there is $a\in A$ such that $1\in I_a$. But then $I_a= R$, contradicting that $I_a$ is a proper ideal of $R$.
Next, we prove that $I$ is an ideal of $R$.
Clearly $I\neq \emptyset$. Let $x,y\in I, r\in R$. Then $x\in I_a$ for some $a\in A$ and $y\in I_b$ for some $b\in A$. Because $I_b$ is an ideal, $ry\in I_b$. Because our collection of ideals is totally ordered, we either have $I_a\subseteq I_b$ or $I_b\subseteq I_a$. Assume wlog the former, i.e. $I_a\subseteq I_b$. Then also $x\in I_b$ and because $I_b$ is an additive subgroup $x+ry\in I_b\subseteq I$.
This proves that $I$ is an ideal.