Proof that $[\overline{\mathbb Q}:\mathbb R\cap\overline{\mathbb Q}]=2$ I have to prove that for the algebric closure of $\mathbb Q$ above $\mathbb C$, $\overline{\mathbb{Q}}$, the dimension of $\mathbb{R}\cap \overline{\mathbb{Q}}$ over $\overline{\mathbb{Q}}$ is equal to $2$.
I have tried to use the dimension theorem, that says: 
$$[\mathbb{C}:\mathbb{R}]=[\mathbb{C}:\overline{\mathbb{Q}}]\cdot [\overline{\mathbb{Q}}:\mathbb{R}\cap\overline{\mathbb{Q}}]\cdot[\mathbb{R}\cap\overline{\mathbb{Q}}:\mathbb{R}]=2$$
but I am not quite sure about it because $\overline{\mathbb{Q}}\subset \mathbb{C}$ and $\mathbb{R}\cap\overline{\mathbb{Q}} \subset \mathbb{R}$ does not imply that
$$[\mathbb{C}:\overline{\mathbb{Q}}]=[\mathbb{R}\cap\overline{\mathbb{Q}}:\mathbb{R}]=1.$$
Am I right?

Hint: Let $z=x+yi \in \mathbb{C}$, with $x,y \in \mathbb R $. Prove that $z \in \overline{\mathbb{Q}} \implies \bar z \in \overline{\mathbb{Q}} \implies x,y \in \mathbb R\cap\overline{\mathbb Q}$.
