Proof

Write $PropIdl(R)_{\subset}$ for the set of proper ideals of $R$, partially ordered by inclusion. We claim that every chain in $PropIdl(R)_{\subset}$ has an upper bound (def.). This then implies the statement by Zorn's lemma (equivalent to the axiom of choice).

To show the claim, assume that $\mathcal{C} \subset PropIdl(R)_{\subset}$ is a chain. We have to produce an $I \in PropIdl(R)$ such that for all $c \in C$ then $c \subset I$.

We claim that such $I$ is provided by the union:

It is clear that if this is indeed a proper ideal, then it is an upper bound of the chain.

To see first of all that this $I$ is an ideal, consider $x_1, x_2 \in I$. There are thus $J_1, J_2 \in \mathcal{C}$ with $x_1 \in J_1$ and $x_2 \in J_2$. Since a chain is total order by definition, either $J_1 \subset J_2$ or $J_2 \subset J_1$. We may assume the former without restriction, otherwise rename $1 \leftrightarrow 2$. Therefore now $x_1, x_2 \in J_2$ and so we may add them there and find that $x_1 + x_2 \in J_2 \subset I$. Similarly if $r \in R$ then $r x_i \in J_2 \subset I$.

Finally to see that this idea $I$ is indeed proper. But since all the $J_i$ are proper, neither of them contains $1 \in R$, and hence $I$ does not contain $1 \in R$.