[step: Existence of a maximal $K$-embedding] Let $\bar{K}$ and $\bar{K}'$ be two algebraic closures of $K$. Consider the set \begin{align*} \mathcal{S} = \{(L,\,\sigma) \mid K \subseteq L \subseteq \bar{K},\; L \text{ a subfield},\; \sigma\colon L \hookrightarrow \bar{K}' \text{ a } K\text{-embedding}\}. \end{align*} This set is nonempty since $(K,\,\iota)$ belongs to $\mathcal{S}$, where $\iota\colon K \hookrightarrow \bar{K}'$ is the inclusion. Partially order $\mathcal{S}$ by $(L_1,\sigma_1) \leq (L_2,\sigma_2)$ when $L_1 \subseteq L_2$ and $\sigma_2|_{L_1} = \sigma_1$. [guided: Zorn's lemma applies] Let $\{(L_i,\sigma_i)\}_{i \in I}$ be a chain in $\mathcal{S}$. Set $L = \bigcup_{i \in I} L_i$ and define $\sigma\colon L \to \bar{K}'$ by $\sigma(\alpha) = \sigma_i(\alpha)$ for any $i$ with $\alpha \in L_i$; this is well-defined because the $\sigma_i$ are compatible on the chain. Then $L$ is a subfield of $\bar{K}$ and $\sigma$ is a $K$-embedding, so $(L,\sigma)$ is an upper bound in $\mathcal{S}$. By Zorn's lemma, $\mathcal{S}$ has a maximal element $(M,\varphi)$. [guided: The maximal field is all of $\bar{K}$] Suppose $M \neq \bar{K}$. Pick $\alpha \in \bar{K} \setminus M$ and let $f \in M[x]$ be its minimal polynomial over $M$. The image $\varphi(f) \in \varphi(M)[x]$ is irreducible in $\varphi(M)[x]$, and since $\bar{K}'$ is algebraically closed it contains a root $\beta$ of $\varphi(f)$. The $M$-isomorphism $M(\alpha) \cong M[x]/(f)$ and the $\varphi(M)$-isomorphism $\varphi(M)(\beta) \cong \varphi(M)[x]/(\varphi(f))$ combine with $\varphi$ to give a $K$-embedding $\tilde{\varphi}\colon M(\alpha) \hookrightarrow \bar{K}'$ extending $\varphi$. This contradicts maximality, so $M = \bar{K}$ and $\varphi\colon \bar{K} \hookrightarrow \bar{K}'$ is a $K$-embedding. [step: The embedding is surjective] It remains to show $\varphi$ is surjective. Take any $\gamma \in \bar{K}'$. Since $\bar{K}'$ is algebraic over $K$, the element $\gamma$ satisfies some polynomial $g \in K[x]$. Because $\varphi$ fixes $K$, the polynomial $g$ also lies in $\varphi(\bar{K})[x]$. Now $\bar{K}$ is algebraically closed, so $g$ splits completely in $\bar{K}[x]$, and applying $\varphi$ shows $g$ splits completely in $\varphi(\bar{K})[x]$. In particular $\gamma$, being a root of $g$ in $\bar{K}'$, must already belong to $\varphi(\bar{K})$. Therefore $\varphi\colon \bar{K} \xrightarrow{\;\sim\;} \bar{K}'$ is a $K$-isomorphism. $\blacksquare$