If $f(x)$ and $f(x)-x$ have only one real root, then $f(f(x))-x$ has only one real root. First edition was: Let $f(x)$ be a polynomial such that $f(x)$ and  $f(x)-x$ have only one real root. How to prove, without derivatives, that 
$f(f(x))-x$ also has only one real root?
Second edition: Let $f(x)=a_3 x^3 + a_2 x^2 + a_1 x + a_0$.  All $a_i \neq 0$. $f(x)$ and $f(x)-x$ have only one real root. How to prove, without derivatives, that 
$f(f(x))-x$ also has only one real root?
 A: More generally, $f(x)=-x^{2n-1}$ gives a counterexample for each positive integer $n$.
A counterexample for the second edition: $f(x)=-x^3-x^2-x-1$.
A: As Jonas Meyer has pointed out, the problem is actually not true.
As suggested, let $f(x)= -x$. Then $f(x) -x = -x -x = -2x$. Considering $f(f(x)) -x= f(-x) -x = -(-x) -x = 0$. That is to say, for $f(x)=-x$, we have shown $f(x) -x$ has only one real root, namely $x=0$, while $f(f(x))-x=0$ has infinitely many real roots.
Response to comments:
Let $f(x)=x^3 - 2 x^2 + 5 x - 1$. Then, $$f(x) -x = x^3 -2x^2 +5x -1 =x^3 -2x^2 +4x -1$$ Solving for $x$ (by Wolfram Alpha) we have that $x \approx 0.284775$ and that there is only one real root. Considering $f(f(x))-x$ we have the following, $$f(f(x))-x = (x^3 -2x^2 +4x -1)^3 - 2(x^3 -2x^2 +4x -1)^2 +4(x^3 -2x^2 +4x -1) -1$$ which can be "simplifed" to, $$f(f(x))-x = x^9-6 x^8+24 x^7-61 x^6+116 x^5-156 x^4+155 x^3-102 x^2+44 x-8$$
Which has only one real root $x \approx 0.379555$. So, you have shown that there does exist a polynomial, namely $f(x) = x^3 -2x^2 +5x -1$ for which it is true. But this does not show in general that your conjecture is true, since there exists at least one counterexample.
@Jonas: I can't think of another counterexample off the top of my head, I don't want to try to hard at thinking about this.
A: Here's a more general way of making counterexamples.  Choose any real numbers $a < b$.  Let $g(x)$ be any polynomial of odd degree whose leading coefficient is positive. Take $f(x) = a + b - x -  \varepsilon (x-a)(x-b) g(x)$, choosing $\varepsilon \ge 0$ small enough that $f$ is decreasing, and thus $f(x)$ and $f(x) - x$ each have only one zero.  But $f(a) = b$ and $f(b)=a$, so $a$ and $b$ are zeros of 
$f(f(x))-x$.  
