I'm having trouble understanding the proof of Proposition 1.1.12 ("Minimum speed") in Fisher & Hasselblatt's Hyperbolic Flows.
Proposition $1.1.12$ ("Minimum speed"). If $\Phi$ is a continuous flow without fixed points on a compact space, then there is a $T_0 > 0$ such that for any $t \in (0, T_0)$ there is a $\gamma_t>0$ with $d(\varphi^t(x),x) \geqslant \gamma_t$ for all $x$.
Proof. If $t$ is such that for all $n \in \mathbb{N}$ there is an $x_n$ with $d(\varphi^t(x_n),x_n) < 1/n$, then an accumulation point $x$ of the $x_n$ satisfies $d(\varphi^t(x),x)=0$. If there is no periodic point, let $T_0:= 1$, otherwise $T_0 := \inf\{t>0 \mid \exists \ x \in X : \varphi^t(x)=x\}>0. \quad \square $
In particular, I'm having trouble connecting the first and second sentences in the proof. I understand why $T_0$ can be chosen arbitrarily when there's no periodic orbit, but I am struggling to prove that $T_0 > 0$ when there's an orbit. I've tried by contradiction, which gives a sequence of $t_n$, $x_n$ for which $\varphi^{t_n}(x_n) = x_n$ and $t_n \to 0$, and extracted a convergent subsequence with a limit point $x$. I feel that $x$ should be an equilibrium point (since it is surrounded by periodic orbits of increasingly small period), which would yield the appropriate contradiction, but I'm having trouble showing this rigorously.
Is this approach on the right track? Is it connected with the first sentence of Hasselblatt's proof?
