The key idea is to use the sublevel sets of $\mathcal{V}$ to confine trajectories. Fix $\epsilon > 0$ small enough that the closed ball $\overline{B}(0, \epsilon) \subseteq E$. Define \begin{align*} \alpha := \min \{ \mathcal{V}(x) : |x| = \epsilon \}. \end{align*} The set $\{|x| = \epsilon\}$ is compact (closed and bounded), and $\mathcal{V}$ is continuous, so the minimum is attained. Since $\mathcal{V}(x) > 0$ for $x \neq 0$, we have $\alpha > 0$. Now define the sublevel set $U_1 := \{x \in E : \mathcal{V}(x) < \alpha\} \cap B(0, \epsilon)$. By continuity of $\mathcal{V}$, $U_1$ is open. Since $\mathcal{V}(0) = 0 < \alpha$, the origin belongs to $U_1$, so there exists $\delta > 0$ such that $B(0, \delta) \subseteq U_1$. For any $x \in B(0, \delta)$, we have $\mathcal{V}(x) < \alpha$. Since $\dot{\mathcal{V}} \leq 0$ along trajectories, $t \mapsto \mathcal{V}(\phi_t(x))$ is non-increasing, so $\mathcal{V}(\phi_t(x)) \leq \mathcal{V}(x) < \alpha$ for all $t \geq 0$. By definition of $\alpha$, the trajectory $\phi_t(x)$ cannot reach the sphere $|x| = \epsilon$. Hence $\phi_t(x) \in B(0, \epsilon)$ for all $t \geq 0$, which is precisely Lyapunov stability.