I have the initial value problem $$\dot x(t) = h(t, x(t)), \quad x(t_0) = x_0$$ where $h$ has the form $$h(t, x)= \begin{cases} h_2(t,x) & g(t,x) \geq 0 \\ h_1(t,x) & \text{otherwise} \\ \end{cases}$$ where $h_1$ and $h_2$ are not necessarily equal. I want to state the necessary conditions on $h$ so that $x(t)$ is continuous on some interval $[t_0,t_1]$. I don't want to use the notion of a caratheodory or filippov solution, so I only want to use the regular existence and uniqueness theorem.
My claim is that sufficient conditions for $x(t)$ to be continuous are that: 1) $h(t,x)$ is piecewise continuous on the interval $[t_0, t_1]$ and 2) $ \partial h(t,x)/ \partial x$ is able to be bounded by a constant. My logic is that you can define the times $\theta_j, \ j = 0, \ldots, m$ such that on each interval $[\theta_j, \theta_{j+1}]$, $h(t,x)$ is continuous except at the endpoints ( and $[\theta_0, \theta_m] = [t_0, t_1]$) . Then starting on the first interval $[\theta_0, \theta_1]$ with the given initial condition, existence and uniqueness gives you $x(t)$ for $\theta_0 \leq t < \theta_1$ (where we assume that some region $\{(t,x): \ldots\}$ can be suitably chosen to guarantee existence on the whole interval). Then on the rest of the intervals $[\theta_{j}, \theta_{j+1}]$, you can take the initial condition to be $$\underset{\epsilon \rightarrow 0}{\lim} x(\theta_{j} - \epsilon)$$ and apply existence and uniqueness to get $x(t), \ \theta_j \leq t < \theta_{j+1}$. Then this shows that $x(t)$ is continuous on the entire interval $[t_0, t_1]$.
Do you think this proof is correct? Am I missing something? I just want $x(t)$ to be continuous on the whole interval.