Having read the classical proof of the existence of an Algebraic Closure (originally due to Artin), I wondered what is wrong with the following simplification (it must be wrong, otherwise why would we bother with the extra complications in Artin's proof?):
Theorem: Let $K$ be a field, then $K$ has an algebraic closure $\bar{K}$ (i.e an algebraic extension that is algebraically closed).
"Proof": Define $A=\{ F \supset K | F \text{ is an algebraic extension of } K\}$ and inherit this with the usual partial order of inclusion. One can check that Zorn's lemma applies (union of a nested chain of algebraic extensions is itself algebraic). Thus take $\overline{K}$ to be a maximal element. It must be algebraically closed for otherwise there is an irreducible polynomial with root in some strictly bigger field. $\blacksquare$
Now here is what I suspect is false about this proof: The definition of $A$ smells like your usual set theory paradoxes like Russell's paradox. In fact one could just as well use the same technique to prove that there exists a "largest set" which of course there does not. I am however under the impression that "most" working mathematicians ignore set theory foundations and just "do" mathematics, so is there a safe way of doing this (i.e: do "concrete everyday mathematics" by avoiding set theory) without getting burnt?