Suppose $F \supset \mathbb{R}$ is an algebraic field extension of degree $\geq 2$.
I would like to show that $F\supset \mathbb{R}$ must be isomorphic either to $\mathbb{R}$ or to $\mathbb{C}\supset \mathbb{R}$. Is it possible to do so without assuming $F$ is already algebraically closed? (We can assume we already know $\mathbb{C}$ is algebraically closed.)
Any algebraic extension of $\Bbb{R}$ is contained in $\Bbb{C}$ as $\Bbb{C}$ is the algebraic closure of $\Bbb{ R}$
$\Bbb{R}\subset F\subset \Bbb{C}$
Then $[F:\Bbb{R}][\Bbb{C}:F]=[\Bbb{C}:\Bbb{R}]=2$
Since $2$ is a prime at least one of the field extension must have degree $1$ i.e either $[F:\Bbb{R}]=1$ or $[\Bbb{C}:F]=1$ .Hence either $F=\Bbb{R} $ or $F=\Bbb{C}$