I'm trying to follow a proof of the above lemma. My (attempted) clean up is as follows, and I will outline where I think we may have an issue.
Let $\phi_t(X)$ denote the through $X$ at time $t$. Let $X^{*}$ be an equilibrium point (for which the Lyapunov function is applied to). Then assume there lies a point $Y$ on a periodic orbit such that $\phi_{0}(Y) \neq X^{*}$. Then given $Y$ lies on a closed orbit, we know that there exists $K>0$ such that for all $n \in \mathbb{N}$, we have: $$\phi_{nK}(Y) = \phi_{0}(Y)$$ Given that a strict Lyapunov function exists, we know that $X^{*}$ is asymptotically stable and so: $$\phi_t(Y) \to X^{*}$$ And so: $$\phi_0(Y) = \lim_{n \to \infty} \phi_{nK}(Y) = \lim_{t \to \infty} \phi_t(Y) = X^{*}$$ Which is a contradiction given that we chose $Y$ such that $\phi_0(Y) \neq X^{*}$ and so there exists no such closed orbit and so there are no limit cycles. $\blacksquare$
The point I'm not sure of is how we can say that $\phi_t(Y) \to X^{*}$, how do we know that the point $Y$ was chosen such that it lies in the basin of attraction of $X^{*}$. If $Y$ was not chosen as such, then doesn't the above proof break down?
The point $Y$ is on a periodic orbit, so that orbit can't converge to $X^*$.
By the way, it seems unnecessary to involve convergence to $X^*$ in the proof. If $\phi_T(Y)=Y$ for some $T>0$, then $L(\phi(T)(Y)) = L(Y)$, which contradicts the fact that a strict Lyapunov function is strictly decreasing along all trajectories except the constant trajectory $X(t)=X^*$. So there can't be any periodic solutions (except $X^*$ if you count that as periodic), limit cycles or not.