Exercise: A known ODE result regarding maximal solutions is the following:
Given a maximal solution $\gamma: (a,b) \rightarrow \mathbb{R}^d$ of $x' = F(t,x)$, where $F$ is a locally lipschitz function defined on an open set $U \subset \mathbb{R}^{1+d}$, then $(t, \gamma(t)) \rightarrow \partial{U}$, when $t \rightarrow a$ or $t \rightarrow b$.
Show that the conclusion of this theorem remains valid for $F$ merely continuous.
Attempt:
Let $F: \mathcal{U} \rightarrow \mathbb{R}^{d}$ be a continuous function and $\gamma:{(a,b)} \rightarrow \mathbb{R}^{d}$ a maximal solution of $x'=F(t,x)$ in a domain $\mathcal{U} \subset \mathbb{R}^{1+d}$.
We know that any continuous function $F: \mathcal{U} \rightarrow \mathbb{R}^{d}$ can be approximated, in a convenient sense, by differentiable functions $F_{n}: \mathcal{U} \rightarrow \mathbb{R}^{d}$. Therefore, we can consider such functions
$$F_{n}: \mathcal{U}{n} \rightarrow \mathbb{R}^{d}$$
of class $C^{\infty}$, where $(\mathcal{U}{n})_{n}$ is an increasing sequence of open sets in $\mathbb{R}^{1+d}$ whose union is $\mathcal{U}$ and the sequence $(F_n)_n$ converges to $F: \mathcal{U} \rightarrow \mathbb{R}^{d}$ uniformly in compacts of $\mathcal{U}$.
For such sequence $(F_n)_n$, consider the respective differential equations $x'=F_n(t,x)$. As the functions $F_n$ are locally lipschtzian on the variable $x$, by Picard's Theorem, for each $n \geq 1$ and given any $(t_0,x_0) \in \mathcal{U}_n$, we can take maximal solutions $$\gamma_n: \mathcal{I}_n \rightarrow \mathbb{R}^{d} $$ of $x'= F_n(t,x)$ satisfying $\gamma _n (t_0)=x_0$, with $t_0 \in \mathcal{I}_n$ and $\mathcal{I}_n= (a_n,b_n) \subset (a,b)$.
Hence, in spite of restricting to a subsequence, if necessary, the sequence $(\gamma_n)_n$ converges to $\overline{\gamma}: (c,d) \rightarrow \mathbb{R}^{d}$.
Now, note that, for each $n \geq 1$, we have $\gamma_n: (a_n,b_n)\rightarrow \mathbb{R}^{d}$ maximal solution of $x'= F_n(t,x)$ in a domain $\mathcal{U}_n$. So we can aply the theorem mentioned previously for each $n \geq 1$, which assures us that
$$(t,\gamma_n(t)) \rightarrow \partial \mathcal{U}_n \hspace{0.1cm} \text{when} \hspace{0.1cm} t \rightarrow a_n \hspace{0.1cm} \text{or} \hspace{0.1cm} t \rightarrow b_n.$$
Also, note that, as $(\mathcal{U}_n)$ is an increasing sequence of open sets, such that $\bigcup_n \mathcal{U}_n = \mathcal{U}$, when $n \rightarrow \infty$, we have $\partial \mathcal{U}_n \rightarrow \partial \mathcal{U}$.
How do I guarantee that $(a_n,b_n) \rightarrow (a,b)$ when $n \rightarrow \infty$? Followed by that, I cannot conclude that $\overline{\gamma}$ is equal to $\gamma$, because, since the function $F$ is only continuous, I don't have unicity of the maximal solution. Is there any construction missing? What can I do to get $\overline{\gamma}$ explicitly? Any help would be appreciated!