If $f(x) \in \mathbb{C}[x]$ let $a$ be a root of $f(x)$ in some extension of $\mathbb{C}$ Let $K$ be a Galois closure of $\mathbb{C}(a)$ over $\mathbb{R}$ and set $G = \operatorname{Gal}(K/\mathbb{R})$ Let $H$ be a Sylow 2-subgroup of $G$ and let $L = K^H$ (the fixed field of $H$ in $K$ . By the Fundamental Theorem of Galois Theory we have $[L:\mathbb{R}] = [G:H]$ an odd number. We may write $L = \mathbb{R}(b)$ for some $b \in L$ so the minimal polynomial $m_{b,\mathbb{R}}(x)$ is irreducible over $\mathbb{R}$ and of odd degree. That degree must be 1, and hence $L = \mathbb{R}$ which means that $G = H$ a 2-group. Thus $G_1 = \operatorname{Gal}(K/\mathbb{C})$ is also a 2-group. If $G_1 \ne 1$ choose $G_2 \le G_1$ such that $[G_1:G_2] = 2$ and set $M = K^{G_2}$ so that $[M:\mathbb{C}] = [G_1:G_2] = 2$ But any polynomial of degree 2 over $\mathbb{C}$ has roots in $\mathbb{C}$ by the quadratic formula, so such a field $M$ cannot exist. This contradiction shows that $G_1 = 1$ Hence $K = \mathbb{C}$ and $a \in \mathbb{C}$ completing the proof.