TITLE: Given a vector field all of whose integral curves are closed, is the period a smooth function?
QUESTION [6 upvotes]: Disclaimer: The original question consisted of two parts. The first one
  has been answered negatively (see
  below the answers of Sam Lisi and
  Alejandro). It remains the second one.

Background
I am reading about the energy-period relation for Hamiltonian Systems.
In Weinstein's formulation (cf. Abraham, Marsden, Foundations of Mechanics 2nd Ed, page 198) this relation amounts to:

$(\ast)$ Given an Hamiltonian system $(M,\omega, X_H)$, let be $\Phi$ the flow of $X_H$ and $\text{per}:=\{(t,x)\in\mathbb R\times M\mid\Phi(t,x)=x\}.$
  If $N$ is a smooth submanifold contained in $\text{per},$ then $\left.dt\wedge dH\right|_N=0,$ i.e. $t=t(H)$ on $N,$ (the period depends only on the energy.)

Question
In Guillemin, Stenberg, Geometric Asymptotics, between pages 170-171, I have additionally found that, when all integral curves of $X_H$ are periodic, we can take $N=\text{per}$ in $(\ast),$ which should mean that in such a case $\text{per}$ is a smooth submanifold of $\mathbb R\times M.$  
In order to justify this last point I was wondering myself:


If $X$ is a non singular vector field on $M,$ all of whose integral curves are periodic, and $\tau(p)$ denotes the period of the orbit through $p,$ then $\tau:M\to\mathbb R$ is smooth? 
otherwise, how to prove that in such a case $\text{per}$ is a submanifold?


What I have tried about point 2
Probably I am missing something because my guess is that if there were a principal bundle structure $(M,p,X,\mathbb S^1)$ such that the $\mathbb S^1$-orbits are the trajectories of $X$ then the period $\tau:M\to\mathbb R$ should be smooth because of the relation $\zeta=\tau X_H,$ where $\zeta$ is the infinitesimal generator of the action.
But I don't know how to proceed without this additional hypothesis.  
Edit1 (After Sebastian's answer about point 1): As illustration of my difficulties with point 1, I imagine that $M$ is the Moebius band $[0,1]\times\mathbb R/\sim$ and $X=\frac{\partial}{\partial x}$ then the period is $$\tau([(x,y)]_{\sim})=\begin{cases}1&\text{if }y=0\\\2&\text{if }y\neq 0\end{cases}$$

REPLY [7 votes]: Sam is completely right. In general, the period function $\tau\colon M\to\mathbb{R}$ is not even continuous. 
A very nice reference for a (counter-)example to Giuseppe's question is the paper A counterexample to the periodic orbit conjecture, by Dennis Sullivan. In the paper, Sullivan constructs a singularity-free flow on a compact 5-manifold such that all its orbits are periodic and function $\tau$ is unbounded!