Question is in the title. Still a little new to Zorn's Lemma, so I wanna make sure what I'm doing checks out. Thank you :)
$\textbf{Proof(existence):}$
Let $Z$ denote all algebraic extensions a field $F$ partially ordered by inclusion. Let $S\subseteq Z$ be a chain. Consider $\bigcup S=S'.$ If $a,b,c\in S',$ then by total ordering there exists $F'\subset S$ with $a,b,c\in F'.$ Since $F'$ is a field it follows that we can add, multiple, and divide any of $a,b,c$, and doing so with these elements respects all the field properties. Then we can add, multiple, and divide any elements of $S'$ and doing so respects all the field properties. Moreover, since $F'/F$ is algebraic $a$ is algebraic over $F,$ so since arbitrary $S'/F$ is algebraic. Then $S'\in Z$ is an upper bound for $Z,$ hence $Z$ has a maximal element, say $K.$
Since $K$ is maximal in $Z$ there exists no $L\in Z$ with $L\supset K\subseteq F,$ hence there is no $L\supset K\supseteq F$ with $L/F$ algebraic, therefore $K$ is an algebraic closure of $F.$ $\blacksquare$
$\textbf{Proof(uniqueness)}:$
Let $F',F''$ be two algebraic closures of $F.$ We let $$Z:=\{h:K\to K'|F\subseteq K\subseteq F',F\subseteq K'\subseteq F'',\text{ }f\text{ is an isomorphism leaving $F$ fixed}\}$$ be indexed by inclusion. Let $S\subseteq Z$ be a chain. Consider $g:=\bigcup S.$ It follows that $g\subseteq F'\times F'',$ we will it is in $Z.$ Let $x\in F'$ and $y,w\in F''$ with $(x,y),(x,w)\in g.$ By total ordering there exists $l\in S$ with $(x,y),(x,w)\in l,$ but $l$ is an isomorphism, so $y=w.$ Then $g$ is a function, and by symmetry it is injective. Trivially $g$ is onto its image. Now suppose $a,b\in\text{Domain}(g),$ then by total ordering and WLOG $x,z\in\text{Domain}(l).$ Since $l$ is a homomorphism $l(ab)=l(a)l(b)$ and $l(a+b)=l(a)+l(b),$ but $g$ is an extension of $l,$ so $g$ is additive and multiplicative as well, hence $g$ is an isomorphism. Clearly $g$ leaves $f$ fixed, as $l$ does. Then $g\in Z$ is an upper bound for $S.$
By Zorn's Lemma $Z$ let $h\in Z$ be maximal. Let $h:K\to K'.$ Now if $K\not= F$, then there exists $m\in F[x]$ irreducible over $K,$ and since $K'$ is an isomorphic image of $K$ m is also irreducible over $K'.$ Then we can extend $h$ to an isomorphic $h':K(\alpha)\to K'(\beta)$ where $\alpha\in F'$ and $\beta\in F''$ are roots of $m$ in there respective algebraic closures. However, $h$ was maximal, so we cannot extend it, and we must have $K=F.$ By symmetry $K'=F''.$ This shows that algebraic closures are unique upto isomorphism. $\blacksquare$
Let $E(F)$ denote the elements of $F$ for any field $F$. $ \def\pow{\mathcal{P}} $
Take any non-algebraically closed field $F$ and any object $x$. Then there exists an algebraic extension $K$ of $F$ such that $x ∈ E(K)$. This is trivial to prove.
If your proof is valid, then we can apply it to $F = ℚ$ and let $U = \{ \ x : ∃K{∈}Z\ ( \ x∈E(K) \ ) \ \}$, where $Z$ is the purported set in your proof. Then $U$ includes every single object; it is the universe, hence a contradiction (via the well-known $R = \{ \ x : x∈U ∧ x∉x \ \}$).
Therefore your proof is invalid. It can be fixed if you do not try to collect all algebraic extensions of $F$ but only those whose elements are taken from some set $T$ that is big enough. One easy choice is $\pow(E(F)×ℕ)$, because for every algebraic extension $K$ of $F$ we have that $E(K)$ does not surject onto $E(F)×ℕ$ and so we always have enough unused elements of $T$ at every step of the extension process. But this easy choice would mean that the Zorn's lemma involved is actually equivalent to existence of a well-ordering of $\pow(E(F)×ℕ)$, which is not optimal. A better solution would be to not order by inclusion but by embedding, in which case it suffices to have $T = E(F)×ℕ$!