Lyapunov's direct method (second method)

Fix a small $\varepsilon>0$ small enough that the closed ball $\{\|x-x^{*}\|\le\varepsilon\}$ lies in the domain where $V$ is defined and positive away from $x^{*}$, and let $S_\varepsilon=\{\|x-x^{*}\|=\varepsilon\}$ be its boundary sphere. Since $V$ is continuous and strictly positive on the compact set $S_\varepsilon$, it attains a minimum $m=\min_{S_\varepsilon}V>0$.

By continuity of $V$ and $V(x^{*})=0$, choose $\delta\in(0,\varepsilon)$ small enough that $\|x(0)-x^{*}\|<\delta$ implies $V(x(0))<m$.

Because $\dot V\le0$ along trajectories, $V(x(t))\le V(x(0))<m$ for every $t\ge0$ that the solution exists. Every point of $S_\varepsilon$ has $V\ge m$, so the trajectory can never reach $S_\varepsilon$; by continuity of $t\mapsto x(t)$, it must remain strictly inside the ball of radius $\varepsilon$ for all $t\ge0$. This is exactly the definition of Lyapunov stability.

If in addition $\dot V(x)<0$ strictly for $x\ne x^{*}$ nearby, then $V(x(t))$ is strictly decreasing and bounded below by $0$, so it converges to some limit $c\ge0$. If $c>0$ were positive, the trajectory would stay in the compact annulus where $V\ge c>0$, on which $\dot V$ attains a strictly negative maximum, forcing $V$ to decrease past $c$ in finite time — a contradiction. Hence $c=0$, and since $V$ is positive-definite this forces $x(t)\to x^{*}$: asymptotic stability.