I credit Mikey and Atal with this proof.
For any set which is a subset of a (lets say much larger) set , a minimal superset of is a set such that , and if , then for any superset of .
I claim that any set has a minimal superset. We see this by using Zorn’s Lemma on the elements of which contain (I denote this collection of sets as ). The ordering will be the reverse of the typical: if such that , then we say . If is a chain in , then this chain is bounded above by . Thus, by Zorn’s Lemma, there is a maximal element .
I claim this is a minimal subset. We see that must contain by definition, and further contains no other elements of . Suppose that is in . We have that and , as both and are in . Further, . Because contains and is a subset of , we are forced to conclude, by maximality of , that , meaning that . Thus, is a minimal superset of .
Of course, this whole proof can be greatly simplified by noting that itself will fulfill the property of being a minimal superset of , so this is a proof by immediate example. Hey, the definition doesn’t forbid it!