Let $v:\left[0,1\right]\to\mathbb{R}$ be a Lipschitz continuous velocity, and let $t\mapsto u(t,x)$ be the (unique) solution of the initial problem
$$x'(t)=v(x(t)),\;\;\; x(0)=x$$
defined on its maximal interval of existence, say $I_x$. Consider the flow $(\Phi_t)_{t\geq 0}$ given by
$$\Phi_t(x)= \begin{cases} u(t,x), &t\in I_x,\\ u(\sup I_x,\,x), &\text{otherwise}. \end{cases} $$ Now define the first hitting time $$\tau_y(x)=\inf\{t\geq 0:\, \Phi_t(x)=y\}\;\;\;(\text{with the convention: }\inf\emptyset:=\infty).$$
I would like to prove that
if $y\in\left[0,1\right]$ satisfies $v(y)=0$ then $\tau_y(x)=\infty$ for all $x\in\left[0,1\right]\backslash\{y\}$.
I have a problem with a proof of this. Let $y\in\left[0,1\right]$, and think about the case where $x<y$. Roughly speaking, I see that the Lipschitz continuity of $v$ forces that $s\mapsto\frac{d}{ds} \Phi_s(x)$ tends to $0$ as $s\mapsto \Phi_s(x)$ gets closer and closer to $y$. However, I have no idea how to formally show that $\Phi_s(x)=y$ cannot happen for some $s>0$. I would be grateful for any hints.
Suppose that $\Phi_s(x)=y$ for $s > 0$. As $v(y) = 0$, $y(t)=y$ is a solution of the initial problem $$x^\prime(t)=v(x(t)),\;\;\; x(s)=y.$$ But $u(t,x)$ is another distinct one. In contradiction with uniqueness of solutions for a Lipschitz continuous differential equation.