This is just to record that the condition $f\in C^2$ in the accepted answer can be weakened to Lipschitz continuity of $f$. With $y=x'$, let us write the equation as $$ y'=f(y). $$ First of all, $y\not\equiv0$ because $f(0)\neq0$. On the other hand, if $f(\eta)=0$ for some $\eta\neq0$ then $y$ cannot take the value $\eta$ because if $y(b)=\eta$ for some $b\in[0,1]$ then by uniqueness we would have $y\equiv\eta$. Now, if $y(t)>0$ for all $t\in(0,1)$ then by definition $x$ would be strictly increasing on $(0,1)$, and likewise $y<0$ on $(0,1)$ implies that $x$ would be strictly decreasing on $(0,1)$, meaning that it would not be possible to satisfy the boundary conditions $x(0)=x(1)=0$. Hence $y$ must change sign on $(0,1)$, i.e., there is $a\in(0,1)$ such that $y(a)=0$. The claim follows from the fact that $y$ is a continuous function that takes the value $0$, but cannot take the values $r_{\pm}$.

This is just to record that the condition $f\in C^2$ in the accepted answer can be weakened to Lipschitz continuity of $f$, and that $r_{\pm}$ do not have to be consecutive zeros. With $y=x'$, let us write the equation as $$ y'=f(y). $$ First of all, $y\not\equiv0$ because $f(0)\neq0$. On the other hand, if $f(\eta)=0$ for some $\eta\neq0$ then $y$ cannot take the value $\eta$ because if $y(b)=\eta$ for some $b\in[0,1]$ then by uniqueness we would have $y\equiv\eta$. Now, if $y(t)>0$ for all $t\in(0,1)$ then by definition $x$ would be strictly increasing on $(0,1)$, and likewise $y<0$ on $(0,1)$ implies that $x$ would be strictly decreasing on $(0,1)$, meaning that it would not be possible to satisfy the boundary conditions $x(0)=x(1)=0$. Hence $y$ must change sign on $(0,1)$, i.e., there is $a\in(0,1)$ such that $y(a)=0$. The claim follows from the fact that $y$ is a continuous function that takes the value $0$, but cannot take the values $r_{\pm}$.