There are open sets $U$ and $V$ such that $U\cap V = \emptyset $ and $ U \cap \tau(V) =\emptyset $?

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 $$U\cap V \subseteq U_1\cap V_1 =\varnothing\quad \textrm{and}\quad U\cap \tau(V)\subseteq U_2\cap \tau(\tau(V_2))=U_2\cap V_2=\varnothing.$$

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 $U \cap \tau(V) \ne \emptyset$.