There are open sets $U$ and $V$ such that $U\cap V = \emptyset $ and $ U \cap \tau(V) =\emptyset $? I am in a series of self-studies in the book Vector Analysis written by Klaus Janich. By page 15 of the book is made the affirmation (without explicit proof) that we can choose (under the considerations below) open sets $U$ and $V$ such that $U\cap V =\emptyset $ and $ U \cap \tau(V) =\emptyset $. 
In this page, $\tau:M\to M$ is a fixed-point-free involution (i.e.
a differentiable map with $\tau\circ\tau=\mathrm{id}_M$ and $\tau(x)\neq x$ for all $x\in M$) and $M$ a $m$-dimensional manifold.

QUESTION. Under these considerations how can we prove that there are open sets $U\neq \emptyset$ and $V\neq \emptyset$ such that $U\cap V = \emptyset $ and $ U \cap \tau(V) =\emptyset $?

In my attempts, the only things I have been able to prove are that
$$
U\cap V=\emptyset\Longleftrightarrow \tau(U)\cap \tau(V)=\emptyset
\qquad \mathrm{ and } \qquad 
U\cap \tau(V)=\emptyset\Longleftrightarrow \tau(U)\cap V=\emptyset
$$
 A: Lee Mosher's answer doesn't quite appear to answer the question asked by the OP, since he finds $U,V$ such that $U\cap V=\varnothing$ but $U\cap \tau(V)\ne \varnothing$ instead of $U\cap \tau(V)=\varnothing$, but a modification of his technique should work.
Pick an arbitrary $x\in M$, and choose $y\ne x,\tau(x)$. Let $V_1,U_1$ be neighborhoods of $x$ and $y$ such that $V_1\cap U_1 =\varnothing$ and let $V_2,U_2$ be neighborhoods of $\tau(x)$ and $y$ such that $V_2\cap U_2=\varnothing$ by Hausdorffness of $M$. Then let 
$$U=U_1\cap U_2 \quad\textrm{and}\quad V=V_1\cap \tau(V_2).$$ $U$ and $V$ are both nonempty, since $U$ contains $y$ and $V$ contains $x$. Then
$$U\cap V \subseteq U_1\cap V_1 =\varnothing\quad \textrm{and}\quad U\cap \tau(V)\subseteq U_2\cap \tau(\tau(V_2))=U_2\cap V_2=\varnothing.$$
A: Pick any $x \in M$. Let $y = \tau(x)$, so we know that $x \ne y$. Since every manifold is a Hausdorff space, there exist open sets $U,V \subset M$ such that $x \in U$, and $y \in V$, and $U \cap V = \emptyset$. Since $\tau(y)=\tau(\tau(x))=x \in U$, it follows that $x \in U \cap \tau(V)$ and so $U \cap \tau(V) \ne \emptyset$.
