Solutions

Let

$$f_m(x)=x^3-3x+m.$$

Suppose now for contradiction that $f$ has two roots $x,y\in[0,1]$ with $x<y$. By Rolle's theorem

$$f'(c)=0$$

for $c$ such that $0\le x<c<y\le 1$. However,

$$\begin{align*} f'(c)=0 &\implies 3c^2-3=0 \\ &\implies c^2=1 \\ &\implies c=\pm 1, \end{align*}$$

which contradicts the supposition that $0<c<1$.