Consider the replicator equation for $2$ strategies:
$$\dot{x} = x(1-x)((a+b)x - b).$$
Prove that for any $x(0) \in [0,1]$, then $x(t) \in [0,1] ~\forall t >0.$
My attempt
Suppose that $x(0) \in (0,1)$ and for a time instant $t$ we have that $x(t) < 0$. Due to the continuity of the solution, there exists another time instant $0 < t' < t$ such that $x(t') = 0$. But $\left.\dot{x}\right|_{x(t') = 0} = 0$. Similarly, suppose for a time instant $t$ we have that $x(t) > 1$. For the continuity of the solution, there exists another time instant $0 < t' < t$ such that $x(t') = 1$. But $\left.\dot{x}\right|_{x(t') = 1} = 0$.
The argument that $\dot{x}_{|_{x=0}}=0$ is not sufficient. Consider for example the equation $$ \dot{x}=-2\sqrt{|x|}. $$ Then $$ x(t) = \left\{ \begin{aligned} (1/2-t)^2, \quad t \leq 1/2, \\ -(1/2-t)^2, \quad t>1/2 \end{aligned}\right. $$ is a solution, $x(0)=1/4>0$ and $\dot{x}_{|_{x=0}}=0$ but $x$ becomes negative for $t>1/2$.
The simplest approach to prove this statement uses uniqueness of solutions. Note that $x\equiv 0$ and $x\equiv 1$ are the solutions to $\dot{x} = x(1-x)((a+b)x - b)$. The solutions to Cauchy problem are unique, therefore any two solutions cannot intersect. In particular, the solution $x(t)$ starting from $x(0)\in(0,1)$ cannot intersect with $x=0$ and $x=1$ and in consequence $x(t)\in (0,1)$ for all $t>0$.