Prove that the union of two ideals is an ideal only if one of the ideals is contained within the other. Here's my proof:

Let the ideal $I = \{i_1, i_2,...\}$ and $J=\{j_1, j_2,...\}$ Then $I \cup J$ is an ideal only if for all $\nu, \mu$, $i_{\nu}+{j_\mu} \in I \cup J$. Let us assume that there exists an element $i_{\nu}$ of $I$ that does not belong to $J$. Then, for any $j_{\mu}$, either $i_{\nu}+{j_\mu} \in J$, which is a contradiction since it would imply that $i_{\nu} \in J$, or $i_{\nu}+{j_\mu} \in I$, which implies that $j_{\mu} \in I$. Letting $\mu$ run through the index of $J$, we get that $J \subset I$. Thus, our proof is complete.

However, here are some things bothering me:


There seems something fishy about this proof, but I can't point it out. Is my proof correct?
Is this a constructive proof or a proof by contradiction?
Does this proof use the Axiom of Choice?


Thanks in advance for any help!
Your proof uses contradiction to show that $j_\mu\in J$.
It does not.


A less fishy proof:

Suppose that there are $i\in I\setminus J$, $j\in J\setminus I$. If $I\cup J$ is an ideal then $i+j\in I\cup J$, but $i+j\in I$ implies $j\in I$ and $i+j\in J$ implies $i\in J$. Both conclusions are contradictions, so there is no such pair $(i,j)$.
This theorem says that if A, B are ideals of ring R and A is subset of B, then the union of A and B is also ideal.
Notice that if A is subset of B, then the union of A and B is B itself. So, the union of A and B which is equal to B is also ideal, since B is ideal.