If the inverse image of a Lyapunov function for a singular point p is contained in a compact set, then it is in the basin of attraction of p

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Let $\Omega\in \mathbb{R}^{N}$ be an open set and $f :\Omega \rightarrow \mathbb{R}^{N}$ be a function with continuous derivative. Furthermore suppose that $p \in \Omega$ is a singular point for the ode $\dot{x}=f(x)$ i.e $f(p)=0$ and $V :\Omega \rightarrow \mathbb{R}$ be a Lyapunov function for p with continuous derivative, and furthermore assume that

(1) the set {$m \in \Omega $ such that $ \dot{V}=0$} has no positive semi trajectory besides the point p

(2) for some $\alpha \gt 0$, ${V}^{-1}[0,\alpha]$ is a compact of $\Omega$

Show that ${V}^{-1}[0,\alpha]$ is contained in the basin of attraction of p.

I found this on the site System with a Lyapunov function over $\mathbb{R}^n$ but not globally asymptotically stable

So we have that the system is globally stable if the Lyapunov function is arbitrarly large if the initial conditions are arbitraly large, otherwise there could be solutions that are in the level sets but are not compact. But since ${V}^{-1}[0,\alpha]$ is compact, it follows from (1) that V is strictly Lyapunov, and hence ${V}^{-1}[0,\alpha]$ is in the basin of attraction.

But it isn`t so clear to me. Can someone refer some textbook or provide an that cold help me to improve my thought with a more insightful idea for this? Or Am I doing some mistake?