Step 1: For each monic non-constant $f \in F[x]$, introduce a variable $x_f$. Form the polynomial ring $S = F[\{x_f : f \in F[x],\, \deg f \geq 1\}]$ and the ideal $I = (f(x_f) : f \in F[x],\, \deg f \geq 1)$.

Step 2: $I \neq S$. If $1 \in I$, then $1 = \sum g_i f_i(x_{f_i})$ for finitely many $f_1, \ldots, f_k$. The product $h = f_1 \cdots f_k$ has a splitting field $K$, and evaluating at roots gives $1 = 0$ in $K$, a contradiction.

Step 3: Since $I \neq S$, extend $I$ to a maximal ideal $\mathfrak{m} \supseteq I$ (by Zorn's lemma). Set $F_1 = S/\mathfrak{m}$, which is a field extension of $F$ in which every $f \in F[x]$ has a root.

Step 4: Iterate: $F \subset F_1 \subset F_2 \subset \cdots$ where each $F_{n+1}$ is constructed from $F_n$ so that every polynomial over $F_n$ has a root in $F_{n+1}$. Set $\overline{F} = \bigcup_{n=0}^{\infty} F_n$.

Every polynomial over $\overline{F}$ has coefficients in some $F_n$, hence has a root in $F_{n+1} \subseteq \overline{F}$. By induction on degree, every polynomial splits completely. So $\overline{F}$ is algebraically closed.

The extension $\overline{F}/F$ is algebraic: every element of $F_n$ is a root of a polynomial over $F_{n-1}$, and by induction, a root of a polynomial over $F$. $\blacksquare$