Let $R$ be a ring with countably many elements, and let $J$ be a proper ideal of $R$.
Without using Zorn's Lemma, show that there must be a maximal proper ideal containing $J$.
How to proceed directly without using Zorn's Lemma or the Axiom of Choice? Does it help to know that every ideal is generated by countably many elements?
The fact the ring is countable means that you can enumerate it as $r_n$ for $n\in\Bbb N$. Now recursively we can define $J_0=J$, and $J_{n+1}$ as the idea generated by $J_n\cup\{r_k\}$ where $k$ is the least element in the index we can add to $J_n$ while preserving it being a proper ideal. Finally, let $I=\bigcup J_n$. It is not hard to see why $I$ is a proper ideal, and by induction we can show that if $r\notin I$, then $I\cup\{r\}$ generate the whole $R$ (otherwise, take the least index of $r$ such that this happens, and find a contradiction).
This sort of thing is exactly what you need to appeal to choice, or to well-ordering, to begin with. In some sense, this is an ad-hoc implementation of the proof of Zorn's lemma from the well-ordering principle, which is apt here, since the claim is classically proved by means of Zorn's lemma, and the countability assumption provides us with a well-order.