An irreducible polynomial in $\mathbb{R}[x]$ has degree $1$ or $2$

By [L1], choose a root $z\in \mathbb{C}$ of $f$. If $z\in \mathbb{R}$, then the minimal polynomial of $z$ over $\mathbb{R}$ is $x-z$, and [L3] makes $x-z$ divide $f$. Since $f$ is irreducible, this forces $\mathrm{deg}f=1$.

Suppose instead that $z\mathrm{\notin }\mathbb{R}$. Then [L2] gives $f(\bar{z})=0$. Put $q(x):=(x-z)(x-\bar{z})=x^2-(z+\bar{z})x+z\bar{z}\in \mathbb{R}[x]\mathrm{.}$ Because $q(z)=0$, fact [L3] makes the minimal polynomial of $z$ over $\mathbb{R}$ divide $q$ and also divide $f$. Since $f$ is irreducible, that minimal polynomial is associated to $f$, so $\mathrm{deg}f\le \mathrm{deg}q=2.$ The degree cannot be $1$ in the nonreal case, hence $\mathrm{deg}f=2$.

The real-root and nonreal-root cases are exhaustive, so every irreducible polynomial in $\mathbb{R}[x]$ has degree $1$ or $2$.