Consider the backwards v.d. Pol oscillator along with some constraint \begin{align} \dot{x}_1 & = - x_2, \\ \dot{x}_2 & = x_1 - (1 - x_1^2)x_2,\\ g(x_1,x_2) & \leq 0, \end{align} where $g(x_1,x_2) = x_1 - 100$.
The (backwards) v.d. Pol oscillator has a well-known limit-cycle. I purposefully chose the constraint so that the limit cycle is contained in the set $\{x\in\mathbb{R}^2 : g(x)\leq 0\}$, where $x \triangleq (x_1,x_2)$.
Let $x^{\bar{x}}(t)$ denote the solution of the backwards v.d. generator at time $t\geq 0$, intiating at $\bar{x}$ at $t = 0$.
Let \begin{equation} \mathcal{R}\triangleq \{\bar{x}: g(x^{\bar{x}}(t)) \leq 0\,\, \forall t\in[0,\infty)\,\,\}. \end{equation} Thus, we have: \begin{equation} \mathcal{R}^\mathsf{C}\triangleq \{\bar{x}: \exists \bar{t} < \infty \text{ s.t. }g(x^{\bar{x}}(\bar{t})) > 0\}. \end{equation}
What is wrong with the following argument?
\begin{equation} \bar{x}\in\mathcal{R} \Leftrightarrow \sup_{t\in[0,\infty[} g(x^{\bar{x}}(t)) \leq 0, \end{equation} \begin{equation} \bar{x}\in\mathcal{R}^\mathsf{C} \Leftrightarrow \sup_{t\in[0,\infty[} g(x^{\bar{x}}(t)) >0. \end{equation} We also have \begin{equation} \bar{x}\in\mathsf{cl}(\mathcal{R}^{\mathsf{C}}) \Leftrightarrow \sup_{t\in[0,\infty[} g(x^{\bar{x}}(t)) \geq 0, \end{equation} and \begin{equation} \partial\mathcal{R} = \mathcal{R}\cap\mathsf{cl}(\mathcal{R}^{\mathsf{C}}), \end{equation} therefore, \begin{equation} \bar{x}\in\partial\mathcal{R} \Leftrightarrow \sup_{t\in[0,\infty[} g(x^{\bar{x}}(t)) \geq 0 \,\,\,\,\text{and} \sup_{t\in[0,\infty[} g(x^{\bar{x}}(t)) \leq 0, \end{equation} thus, \begin{equation} \bar{x}\in\partial\mathcal{R} \Leftrightarrow \sup_{t\in[0,\infty[} g(x^{\bar{x}}(t)) = 0. \end{equation} But this is nonsense! $\partial\mathcal{R}$ is the limit cycle, and I've placed the constraint faaaar away, so $\bar{x}\in\partial\mathcal{R} \Rightarrow\sup_{t\in[0,\infty[} g(x^{\bar{x}}(t)) < 0$.
I do not know much about the v.d. Pol oscillator, but I assume that the solution $x^{\overline{x}}$ will depend continously (in a suitable sense) on $\overline{x}$. It is then plausible that
$$\bar{x}\in \mathsf{cl}(\mathcal{R}^{\mathsf{C}}) \Rightarrow \sup_{t\in[0,\infty[} g(x^{\bar{x}}(t)) \geq 0 .$$
But I do not see why the converse should be true. If you have a solution such that $\sup_{t\in[0,\infty[} g(x^{\bar{x}}(t)) = 0$, why should it be the limit of solutions in $\mathcal{R}^{\mathsf{C}}$?
You have the constraint $g(x_1,x_2) \le 0$ so that $\mathcal{R}^{\mathsf{C}} = \emptyset$ and $\mathsf{cl}(\mathcal{R}^{\mathsf{C}}) = \emptyset$. In your example the zero function is a solution initiating at $\overline{x} = 0$, but is not in $\mathsf{cl}(\mathcal{R}^{\mathsf{C}})$.