Let's first state clearly the question you have.
Let $F$ be a field. An "algebraic closure" of $F$ is a field extension $C/F$ such that
- $C$ is algebraic over $F$;
- $C$ is algebraically closed.
Claim:
If $C_1/F, C_2/F$ are two algebraic closures of $F$, then there exists an $F$-isomorphism $\iota: C_1 \simeq C_2$.
Remark: This is not a universal property, as we don't require (and it is usually impossible) that the isomorphism $\tau$ be unique.
Proof: We first use Zorn's lemma to show that there is an $F$-embedding (i.e. a homomorphism of fields which is identity on $F$) from $C_1$ to $C_2$.
Let $A$ be the set of pairs $(K, \iota)$, where:
- $K/F$ is a subextension of $C_1/F$;
- $\iota: K \rightarrow C_2$ is an $F$-embedding.
The set $A$ is non-empty, as it contains the element $(F, id)$ where $id$ stands for the canonical embedding of $F$ into $C_2$.
A partial order $\leq$ is defined on $A$ as follows. We say that $(K, \iota) \leq (K', \iota')$ if $K$ is a subextension of $K'$ and $\iota'|_K = \iota$.
Now if we have a chain $(K_i, \iota_i)_{i \in I}$ in $A$, we build an element $(K, \iota)$ with $K = \bigcup_{i\in I}K_i$ and $\iota: K \rightarrow C_2$ defined by $\iota(x) = \iota_i(x)$ for any $i\in I$ and any $x\in K_i$, which is well defined because of the chain condition.
This is then an upper bound of the chain.
Therefore, our set $A$ satisfies the condition in Zorn's lemma, and consequently $A$ contains at least one maximal element, say $(K, \iota)$.
Suppose that $K \neq C_1$. Then there exists $x \in C_1 \backslash K$. We let $K'$ be the field $K(x)$.
Since $C_1$ is algebraic over $F$ and hence algebraic over $K$, we may take the minimal polynomial $f$ of $x$ over $K$. Now $f$ has at least one root $y$ in $C_2$, and from results of rupture fields (see below *), we see that $\iota$ extends to a homomorphism $\iota': K' \rightarrow C_2$ such that $\iota'|_K = \iota$ and $\iota(x) = y$.
This means that we have $(K, \iota)$ is strictly smaller than $(K', \iota')$, which contradicts the maximality of $(K, \iota)$.
We have thus proved that $K = C_1$ and hence the existence of an $F$-embedding $\iota: C_1\simeq C_2$.
It only remains to see that $\iota$ is surjective.
For any $y\in C_2$, let $g\in F[X]$ be the minimal polynomial of $y$ over $F$ (or any polynomial in $F[X]$ such that $g(y) = 0$).
Since $C_1$ is algebraically closed, $g$ decomposes in $C_1$ as a product of linear factors: $g = \prod_{i = 1}^d(X - x_i)$.
Taking images under $\iota$, we get an identity $g = \prod_{i = 1}^d(X - \iota(x_i))$ in $C_2[X]$.
Putting $X = y$ gives $0 = g(y) = \prod_{i = 1}^d(y - \iota(x_i))$ as an identity in $C_2$.
But $C_2$ is a field, hence $y$ must be equal to one of the $\iota(x_i)$.
Thus we have shown that $\iota$ is surjective.
(*) In fact, we have $K' = K(x) \simeq K[X]/(f)$, so we may first define a ring homomorphism $K[X]\rightarrow C_2$ sending $X$ to $y$. Since $f(y) = 0$, the ring homomorphism passes to quotient and gives $\iota'$.
Best Answer
You are too generous with equality and you start by assuming a not yet specifically defined $\overline K$.
I'd suggest you start with
Lemma. Let $K$ be a field, $L/K$ algebraic (but not necessarily algebraically closed), $M/K$ algebraically closed (but not necessarily algebraic). Then there exists at least one field homomorphism $f\colon L\to M$.
Proof sketch. Consider the set of all field homomorphisms $\phi\colon L'\to M$ where $K\subseteq L'\subseteq L$ and define a partial order on this set by saying $\phi\le \psi$ if $\psi$ is an extension of $\phi$ (i.e., the domain of $\phi$ is a subfield of the domain of $\psi$ and $\phi$ is the restriction of $\psi$ to that subfield). Verify that Zorn's lemma can be applied. Pick a maximal element $\phi_\max\colon L_\max\to M$ and show that in fact $L_\max=L$ (using that $L$ is algebraic and $M$ is algebraically closed).
Once you have that, apply the lemma to the case that $L$ and $M$ are both algebraic and algebraically closed and show that any $f$ the lemma gives you is in fact onto.