Let $f:\mathbb{R}^n \to \mathbb{R}^n $ be a continuously differentiable function. Prove that two distinct solutions of $x'(t)=f(x(t))$ (the ODE is autonomous) cannot intersect at any point, not even at different times.
Notes: Clearly there is a relationship between this question and the theorems for the existence and uniqueness of ODEs.
Theorem: Let A be an open subset of $\mathbb{R}^n$ and that $f \in C^1(A)$. Then $\forall x_0 \in A \ \exists \ \alpha>0$ such that the IVP $$x'(t)=f(x(t)), \ x(0)=x_0$$ has a unique solution $x(t)=x(t,x_0)$ on the interval $[-\alpha,\alpha]$.
As $f \in C^1$ the theorem applies so the solutions to $x'(t)=f(x(t))$ are unique (they will not be in the same position at the same time). However I do not know how to show that the two solutions cannot intersect at later times.
Assume there are two different solutions $y_1$ and $y_2$ of the ODE. Then we have points $a$ and $b$ with $y_1(a)=y_2(a)$ and $y_1(b) \neq y_2 (b)$. Therefore the set $$ M:= \left\{x \in [a,b] \mid y_1(x)=y_2(x) \right\} \subset [a,b] $$ is nonempty and bounded. Hence $s= \sup M$ exists and we can find a sequence $(x_n)_n$ with $x_n \to s$. $y_1$ and $y_2$ are continuous and therefore $$ 0 = \lim_n \left( y_1(x_n) - y_2(x_n)\right)=y_1(s)-y_2(s).$$ This shows $c:=y_1(s)=y_2(s)$. Further we have $s<b$ and $$y_1(x)\neq y_2(x) \qquad \text{for all} \ \ x\in (s,b]. \qquad (\star)$$ We have finally shown that $y_1$ and $y_2$ are solutions of $$y'= f(y), \qquad y(s)=c.$$ Since this ODE has only unique solution we get $y_1 =y_2$ in a small neighborhood of $s$. This is a contradiction to $(\star)$.