Since the $\omega$-limit is invariant, then it must contain only points with $\dot{V}(x)=0$ I am trying to understand the following proof:

I do not fully get why he says: since $\Gamma^+$ (which I think is usually referred to as the $\omega$-limit) is an invariant set, then $\dot{V}(x)=0$ in $\Gamma^+$.
Intuitively, it seems to me that, since $\omega$-limit is invariant, it must be composed of singularities or closed orbits (and I know that the $\omega$-limit is connected because the orbits for positive times are trapped inside a compact set). But closed orbits are impossible because $V$ is strictly decreasing, so $\omega(x)$ must only contain singularities.
But even if this reasoning is correct, it seems to me that it is not rigorous as it is. 

The preceding sentence says that $V(p)=c$ for each $p \in \Gamma^+$. And since $\Gamma^+$ is invariant, any orbit $y(t)$ which starts in $\Gamma^+$ stays in $\Gamma^+$. Thus $V(y(t))=c$ (identically) for such an orbit, which implies $\frac{d}{dt} \Bigl( V(y(t)) \Bigr)=0$. And this is exactly what the phrase “$\dot V=0$ on $\Gamma^+$” means.
