Here's my attempt:
Suppose that $p(x)=a_0 + a_1x + a_2 x^2 + \ldots + a_n x^n$ where $n$ is odd, $a_i$ are constants with $a_n \ne 0$. Assume that $a_n > 0$. Then $$ p(x)=a_0 + a_1x + a_2 x^2 + \ldots + a_n x^n = x^n \left( \frac{a_0}{x^n} + \ldots + a_n \right) $$
Now as $x \to -\infty $, $x^n \to - \infty$ and $\left( \frac{a_0}{x^n} + \ldots + a_n \right) \to a_n > 0$, and so $p(x) \to -\infty $. And similarly as $x\to +\infty $ , $p(x) \to +\infty$. By definition of $\lim_{x \to \pm \infty } p(x) = \pm \infty$, for any $M>0$ there are points $x_1$ and $x_2$ such that $p(x_1) < -M < 0 < M < p(x_2)$ and thus from the intermediate value theorem, there is a point c between $x_1$ and $x_2$ with $p(c)=0$
If $a_n <0$, we can do the same as above.
Is my proof correct? And I was wondering how would I write my proof in a single case setting since IVT is applied in two very similar cases ($a_n > 0$ and $a_n <0$). Hints would be enough.
It's good, but there is no need for that, in my opinion: I suppose you're familiar with the fact that if $p(x)\in\mathbb{R}[x]$ then $p(z)=0\iff p(\overline{z})=0$ and that the multiplicities of $z,\overline{z}$ are the same. Since $p(x)$ has exactly as many roots (counting multiplicities) as its degree, the oddness of $\deg p(x)$ implies that there must be a root $w$ such that $w=\overline{w}$. Indeed: otherwise, if $z_1,\overline{z_1},\dots, z_n,\overline{z_n}$ were its roots, it would be $\deg p(x)= 2(M(z_1)+\dots+M(z_n))$, an even number.