I need to prove that if $A$ is a local ring, then the set of non-units form an ideal. To do this it is suggested that I should use Zorn's lemma, but I seem to have found a way to do without. In particular, we consider the suggestive notation $I:=A\setminus A^\times$, then suppose there exists an ideal $J$ of $A$ with $$\langle I\rangle\subsetneq J\subseteq A$$ Then, since $I\subseteq\langle I\rangle\subsetneq J$, the ideal $J$ must contain a unit, so $J=A$, meaning $\langle I\rangle$ is maximal in $A$. Since $A$ is local with unique maximal ideal $\mathfrak{m}$, we must have that $\mathfrak{m}=\langle I\rangle$. Then, if $m\in\mathfrak{m}$, since $\mathfrak{m}\neq A$, $m$ must be a non-unit, and so $m\in I$. Then, for $i\in I$, we clearly have $i\in\langle i\rangle\subseteq\langle I\rangle$, so $i\in\mathfrak{m}$. These combine to give $I=\mathfrak{m}$, which is itself an ideal, meaning $I=A\setminus A^\times$ is an ideal of $A$.
Is my proof correct? Am I implicitly using Zorn's lemma here without realising?
Your proof is incomplete since you have not shown that $\langle I\rangle$ actually is a proper ideal (it could be all of $A$, in which case it would not be a maximal ideal).
In fact, this cannot be proved without the axiom of choice (assuming you define "local ring" as "commutative ring with exactly one maximal ideal"). If you do not assume the axiom of choice, there can consistently exist a nonzero commutative ring $B$ with no maximal ideals. Then, if you take a field $k$, the product ring $A=k\times B$ will have exactly one maximal ideal, namely $0\times B$. However, this maximal ideal does not contain all the non-units of $A$ (specifically, it does not contain $(1,0)$). And the set of non-units of $A$ is not an ideal, since $(1,0)$ and $(0,1)$ are both non-units but their sum $(1,1)$ is a unit.