Fix $x \in D$ and define $c := \inf_{t \geq 0} \mathcal{V}(\phi_t(x))$. Since $D$ is compact and forward-invariant, the orbit $\mathcal{O}^+(x)$ is bounded. By the properties of $\omega$-limit sets established in Chapter 1, $\omega(x)$ is non-empty. Let $y \in \omega(x)$. Then there exists a sequence $t_n \to \infty$ with $\phi_{t_n}(x) \to y$. Since $\dot{\mathcal{V}} \leq 0$ along trajectories, $t \mapsto \mathcal{V}(\phi_t(x))$ is non-increasing and bounded below by $0$, hence convergent. By continuity of $\mathcal{V}$, $\mathcal{V}(y) = \lim_{n \to \infty} \mathcal{V}(\phi_{t_n}(x)) = c$. It remains to show $y \in M_c$, i.e., $\mathcal{V}(\phi_s(y)) = c$ for all $s \geq 0$. Suppose for contradiction that $\mathcal{V}(\phi_s(y)) < c$ for some $s > 0$. By continuity of the flow and of $\mathcal{V}$, there exists a neighbourhood $U$ of $y$ such that $\mathcal{V}(\phi_s(z)) < c$ for all $z \in U$. Since $\phi_{t_n}(x) \to y$, we have $\phi_{t_n}(x) \in U$ for all sufficiently large $n$, so $\mathcal{V}(\phi_{t_n + s}(x)) < c$. This contradicts $c = \inf_{t \geq 0} \mathcal{V}(\phi_t(x))$. Hence $\mathcal{V}(\phi_s(y)) = c$ for all $s \geq 0$, confirming $y \in M_c$.