Field extension of $\R$ of odd degree

Let $x$ be an element of $E$, and let’s show that $x\in \R$. Consider the field $\R(x)$. We have the tower of extensions:
$$\R \subseteq \R (x) \subseteq E.$$By the Tower Law, we have that:
$$[E:\R]=[\R(x):\R][E:\R(x)]$$but since $\R \subseteq E$ is supposed to be of odd degree, by definition, $[E:\R]$ is odd, so both factors $[\R(x):\R]$ and $[E:\R(x)]$ are odd.
Now let $P$ be the minimal polynomial of $x$ over $\R$. Its degree is equal to $[\R(x):\R]$. In other words, $P$ is a polynomial of odd degree. According to a corollary of the Intermediate Value Theorem, this means that $P$ admits a real root $\alpha$, hence it is divisible by $(X-\alpha)$ in $\R$. But it was supposed to be irreducible. The only option is that $P(X)=X-\alpha$, and so $x=\alpha$, meaning that $x$ is a real number.