@Gerry: Oh, very true. I didn't think of that.
The proof is easy, I think. Let $f$ be irreducible mod $m$ of degree $n$. Without loss of generality, let $f$ be monic. If $n$ is odd, $f$ has a real root. If $n$ is even, $M := \text{min}_{x \in \mathbb{R}} f(x) > -\infty$. If M < 0, $f$ has a root. If $M>0$, define $g(x) := f(x) - m\lceil\frac{M}{m}\rceil$. Then $f \equiv g$ modulo $m$ and $g$ has a root $\zeta \in \mathbb{R}$ (by means of the intermediate value theorem).
Edit: You can also find a polynomial that splits over $\mathbb{R}$.
Let $\widetilde{g}$ be a polynomial such that $g(x) = \widetilde{g}(x) \cdot ( x - \zeta )$. We can repeat the same procedure with $\widetilde{g}$, but $\text{deg}(\widetilde{g}) = \text{deg}(g)-1$. By induction we are done.
(The statement we proved by induction on $n$ is as follows: Let $f$ be a polynomial of degree $n$ irreducible modulo $m$. Then there exists a polynomial $g \equiv f$ modulo $m$ such that $g$ splits over $\mathbb{R}$.)