Lyapunov's First Theorem gives Lyapunov stability and confines all trajectories starting in $B(0, \delta)$ to the region $U_1 \subseteq B(0, \epsilon)$. It remains to show that $\phi_t(x) \to 0$ as $t \to \infty$. Fix $x_0 \in B(0, \delta) \setminus \{0\}$. Since $\mathcal{V}(x_0) > 0$ and $\dot{\mathcal{V}} < 0$ along the trajectory, $t \mapsto \mathcal{V}(\phi_t(x_0))$ is strictly decreasing and bounded below by $0$. Thus the limit $\alpha := \lim_{t \to \infty} \mathcal{V}(\phi_t(x_0)) \geq 0$ exists. Suppose for contradiction that $\alpha > 0$. Then $\mathcal{V}(\phi_t(x_0)) \geq \alpha$ for all $t \geq 0$, which by continuity of $\mathcal{V}$ implies that $|\phi_t(x_0)|$ is bounded away from zero. The trajectory therefore lies in the closed annulus $A_{\alpha, \epsilon} := \{x : \mathcal{V}(x) \geq \alpha\} \cap \overline{B}(0, \epsilon)$, which is compact. Since $\dot{\mathcal{V}} < 0$ on $A_{\alpha, \epsilon}$ (which excludes the origin) and $A_{\alpha, \epsilon}$ is compact, there exists $b > 0$ such that $\dot{\mathcal{V}}(x) \leq -b < 0$ for all $x \in A_{\alpha,\epsilon}$. Integrating gives $\mathcal{V}(\phi_t(x_0)) \leq \mathcal{V}(x_0) - bt$, which becomes negative for $t > \mathcal{V}(x_0)/b$. This contradicts $\mathcal{V} \geq 0$. Hence $\alpha = 0$, i.e., $\mathcal{V}(\phi_t(x_0)) \to 0$. By continuity of $\mathcal{V}$ and positive definiteness, $|\phi_t(x_0)| \to 0$.