Say $f(x) \in \mathbb{R}[x]$ has degree $n \geq 3$. If $n$ is odd, then $f(x) \to + \infty$ as $x \to +\infty$ and $f(x) \to – \infty $ as $x \to -\infty$ (or the other way around), so $\exists c \in \mathbb{R}$ such that $f(c) = 0$ by the intermediate value theorem, so $f(x) = (x – c)g(x)$ for a nonconstant polynomial $g(x) \in \mathbb{R}[x]$, so $f(x)$ is reducible.
The same is true if $n$ is even: $f(x)$ is still reducible. Can we prove this without using the fact that $\mathbb{C}$ is algebraically closed? Is there any intuitive reason that an even-degree polynomial in $\mathbb{R}[x]$ is reducible if it is not quadratic? The preceding paragraph gives a simple explanation for why odd-degree polynomials behave this way.
Here is a proof using algebraic closure. First, we must prove that $\mathbb{C} \cong \mathbb{R}[x]/(x^2 + 1)$ is indeed algebraically closed. Let $f \in \mathbb{C}[x] – \mathbb{C}$, and take $M$ to be the splitting field of $f$ over $\mathbb{C}$, then take $N$ to be a normal extension of $\mathbb{R}$ containing $M$, so that $\mathbb{R} \subset \mathbb{C} \subseteq M \subseteq N$; it suffices to prove that $N = \mathbb{C}$. Let $H$ be a Sylow 2-subgroup of $\mathrm{Gal}(N/\mathbb{R})$, so that $[\mathrm{Gal}(N/\mathbb{R}) : H] = [N^H : \mathbb{R}]$ is odd — in fact, $[N^H : \mathbb{R}] = 1$ by the first paragraph of this post, so $H = \mathrm{Gal}(N/\mathbb{R})$, so $[N : \mathbb{C}]$ is a power of $2$ since it divides $|H|$. If $N \neq \mathbb{C}$, then there is a degree-$2$ extension of $\mathbb{C}$, but no such extension can exist because every element of $\mathbb{C}$ has a square root. Therefore $N = \mathbb{C}$ as desired.
Now suppose there is an irreducible, monic polynomial $f(x) \in \mathbb{R}[x]$. As $\mathbb{C}$ is algebraically closed, $\mathbb{C}$ contains a root $\alpha$ of $f$, so $[\mathbb{R}(\alpha) : \mathbb{R}] \leq [\mathbb{C} : \mathbb{R}] = 2$. Also $\mathbb{R}(\alpha) \cong \mathbb{R}[x]/(f)$ via a field isomorphism that is the identity on $\mathbb{R}$, so
$$ \deg f = [\mathbb{R}[x]/(f) : \mathbb{R}] = [\mathbb{R}(\alpha) : \mathbb{R}] \leq 2 $$
Best Answer
You don't need algebraic closedness of $\mathbb{C}.$ All you need is that every polynomial has a complex root (this is proved in a particularly painful way here). If the root is real, you are cooking with gas, if it is complex, it's complex conjugate is also a root, so $(x-r)(r-\overline{r})$ is a degree two factor with real coefficients. Of course, since the first sentence also proves the fundamental theorem of algebra, I am not sure what the upside is. I suppose the real question is: can one prove the statement without using complex numbers at all. A couple of even more horrible ways of showing this were suggested in the answers to this: Proving (without using complex numbers) that a real polynomial has a quadratic factor