Lyapunov's Stability Theorem: Let $x = 0$ be an equilibrium point for the autonomous system $$\dot{x}(t) = f( x(t) ),$$ and $D \subset \mathbb{R} ^ n$ be a domain containing the equilibrium point, i.e., $x = 0 \in D$. Let $V : D \to \mathbb{R}$ be a continuously differentiable function such that $$V (0) = 0, \; \text{and} \; V (x) >0 \; \text{in} \; D \backslash \{0\},$$ $$\dot{V} \leq 0 \; \text{in} \; D.$$
Then, $x=0$ is stable. If, in addition, $$\dot{V}(x)<0 \; \text{in} \; D \backslash\{0\}$$ then $x=0$ is asymptotically stable.
I am trying to show that $x=0$ is asymptotically stable; specifically I would like to show that $$\forall \varepsilon > 0, \; \exists\, T > 0 \; \mathrm{s.t.} \; t > T \implies || x ( t ) - 0 || < \varepsilon.$$
The intuitive sketch of the proof is that one has to fit a sublevel set of continuous functions $V : D \to \mathbb{R}$ inside a ball of radius $\varepsilon$ and show penetration of the solution. In other words, contain a ball inside a level set and use the facts of convergence and continuity of $V$, to show that $V$ is entirely contained in that ball.
We always have $\Omega_b \subset B_{\varepsilon}(0)$ (which I think is a consequence of continuity of $V$).
By continuity of $V$, there exists $d>0$ such that $||x-0||<d\implies V(x)<b$.
If we consider $$\min V(x) > b,$$ where $x \, \in \, \partial B_{\varepsilon}(0)$ and the boundary of the open ball is compact, then
how can we show that $$||x(t)-0||<\varepsilon \; ?$$
PS I am aware that for asymptotic stability, it is sufficient to show that $V(x(t)) \to 0$ as $t \to 0$ but I am more interested to see how $$V(x(t)) \to 0 \; \text{as} \; t \to \infty \implies x(t) \to 0 \; \text{as} \; t \to \infty$$
I'd appreciate any help. Thank you.