Is it true that a 3rd order polynomial must have at least one real root?

It is true that a cubic polynomial must have a real root. Since the lead coefficient is not $0$, we have that $$ \lim_{x\to-\infty}ax^3+bx^2+cx+d=\left\{\begin{array}{}-\infty&\text{if }a>0\\+\infty&\text{if }a<0\end{array}\right. $$ and $$ \lim_{x\to+\infty}ax^3+bx^2+cx+d=\left\{\begin{array}{}+\infty&\text{if }a>0\\-\infty&\text{if }a<0\end{array}\right. $$ Since a polynomial is continuous, by the Intermediate Value Theorem, if it takes a positive value and a negative value, it must take every value in between, in particular $0$.

If $p(x) = ax^3 + bx^2 + cx + d $ and if $a$ and $d$ happen to be of same sign, then you can conclude that there must be at least one negative root. Probably this is what your friend would have done as well.

EDIT

The question was changed from having a negative root to there is at least one real root.

This is a consequence of the fact that $p(x)$ is continuous and the intermediate value theorem for continuous functions.

The important observation is that $$\lim_{x \rightarrow \infty} p(x) = \text{sign(a)} \cdot \infty$$ and $$\lim_{x \rightarrow -\infty} p(x) = -\text{sign(a)} \cdot \infty$$

For instance, if $a > 0$, then $\displaystyle \lim_{x \rightarrow \infty} p(x) = \infty$ and $\displaystyle \lim_{x \rightarrow -\infty} p(x) = -\infty$.

Since the function is continuous, by intermediate value theorem, it must hit all values between $- \infty$ and $\infty$ for some $x$. Hence, in particular, it must hit $0$. Intutively, the function is positive for some large $x$ and negative for some large $x$ and since the function has no jumps, it must hit $0$ somewhere on the real axis before it changes sign.

Any polynomial $p(x)$ of odd degree satisfies $$ \lim_{x \rightarrow \infty} p(x) = \pm \infty $$ and $$ \lim_{x \rightarrow -\infty} p(x) = \mp \infty. $$ That is, one side of the graph goes up forever and the other side of the graph goes down forever. Since polynomials are also continuous, the graph has no choice but to cross the $x$ axis somewhere, giving a real root. (The fact that the continuous function has "no choice" is formalized by the Intermediate Value Theorem).