Every embedding of a splitting field into its algebraic closure, is an automorphism

I would like verification of my proof of the above claim. That is, I am trying to prove that if $$K\subseteq\overline{F}$$ is a splitting field for $$\{f_i\}_I$$ over $$F,$$ where $$\overline{F}$$ is an algebraic closure, then given $$\sigma:K\to\overline{F}$$ an isomorphism leaving $$F$$ fixed, it follows that $$\sigma$$ is an automorphism. The proof I am trying to verify uses Zorn's lemma. A lemma which is quickly becoming a favorite of mine. The part where it says 'previous result' is quoting a result that was done by my professor, which said that if $$\sigma:E\to E$$ is an embedding leaving $$F$$ fixed, then it is actually an automorphism. This result holds for any extension $$E$$ of $$F$$

$$\textbf{Proof:}$$

Let $$K$$ be a splitting field for $$\{f_i\}_I\subseteq F[x]$$ over $$F$$. Let $$\overline{F}$$ be an algebraic closure for $$F.$$

let $$Z=\{K'\subseteq K:\text{ every isomorphism of }K'\text{ onto a subfield of }\overline{F}\text{ leaving}$$ $$F\text{ fixed is an automorphism}\}.$$ We index as usual by inclusion, and assume that $$S\subseteq Z$$ is a chain. Now we consider $$\bigcup S.$$ If $$a,b,c\in\bigcup S,$$ then there exists a field $$E\in S$$ with $$a,b,c\in E$$, so all our field properties hold for $$a,b,c$$ in $$E,$$ and hence in $$\bigcup S,$$ so $$\bigcup S$$ is a field. Now let $$\tau$$ be an embedding of $$\bigcup S$$ into $$\overline{F}$$ leaving $$F$$ fixed. Then $$\tau|_E$$ an an automorphism of $$E$$ as $$E\in Z,$$ therefore $$\tau(a)\in E\subseteq\bigcup S,$$ hence since $$a$$ was arbitrary $$\tau$$ is actually an embedding of $$\bigcup S$$ into $$\bigcup S.$$ Then by a previous result it follows that since $$\tau$$ leaves $$F$$ fixed that $$\tau$$ is actually an automoprhism. Thus every embedding of $$\bigcup S$$ into $$\overline{F}$$ leaving $$F$$ fixed is an automorphism, hence $$\bigcup S\subseteq Z$$ is an upper bound for $$S.$$ By Zorn's Lemma let $$M\in Z$$ be maximal. We wish to show that $$M=K.$$

Suppose $$M\not= K.$$ Since $$K$$ is a smallest subfield of $$\overline{F}$$ such that $$\{f_i\}_I$$ all split in $$K$$ it follows that some $$f_i$$ must not split in $$M.$$ Let $$f_i=c(x-\beta_1)...(x-\beta_m).$$ Now we consider $$M'=M(\beta_1,...,\beta_m).$$ Let $$\rho$$ be an embedding of $$M'$$ into $$\overline{F}.$$ Since $$M\in Z$$ we find that $$\rho|_M:M\to M.$$ As products of powers of $$\beta_1,...,\beta_m$$ form a basis for $$M'$$ over $$M$$ is we show $$\rho(\beta_j)\in \{\beta_1,...,\beta_m\}$$ for all $$j,$$ then we'll know $$\rho$$ is an into map. As $$\rho(\beta_j)$$ is a root of $$\rho(f_i)=f_i$$ (F is fixed). So $$\rho(\beta_i)\in\{\text{roots of }f_i\}=\{\beta_1,...,\beta_m\}.$$ it follows that $$\rho$$ is an embedding from $$M'$$ into $$M'$$ leaving $$F$$ fixed, hence is an automorphism. Well this contradicts the maximality of $$M,$$ therefore $$M=K.$$ This completes the proof. $$\blacksquare$$