I was reading the lemma 4.1 in "J.M.Lee - Introduction to smooth manifolds" which says that given a smooth $n$-manifold $M$, then the tangent bundle $TM$ is a smooth $2n$-manifold.
If $\pi: TM\rightarrow M$ is the natural projection, given an atlas $\mathcal A=\{(U_i,\phi_i)\}$ on $M$, we can define a collection of sets $\{\pi^{-1}(U_i)\}$ and a collection of functions $\widetilde\phi_i:\pi^{-1}(U_i)\rightarrow \mathbb R^{2n}$ which both satisfy the lemma 1.23 namely the construcion lemma for smooth manifolds.
More precisely, on $TM$ is defined the topology having as base the set
$B=\{\widetilde{\phi_i}^{-1}(V)\; \textrm{for all $i$}:\,\textrm{$V$ is open in $\mathbb R^{2n}$}\}$
I have some problems to prove that $TM$ is Hausdorff: clearly it is enough to show that given two distinct point $P=(p,X),Q=(q,Y)\in TM$, either then exist some $\pi^{-1}(U_i)$ containing both $P$ and $Q$ or there exist disjoint sets $\pi^{-1}(U_i)$ and $\pi^{-1}(U_j)$ with $P\in \pi^{-1}(U_i)$ and $Q\in \pi^{-1}(U_j)$. If $P$ and $Q$ lie in the same fiber of $\pi$ it is all clear, but if they lie in different fibres than $\pi (P)=p\neq q=\pi(Q)$ and J.M. Lee says
there exist disjoint smooth coordinate domains $U,V$ for $M$ such that $p\in U$ and $q\in V$
Why is this true? $M$ is Hausdorff, but open sets of $M$ are not the smooth coordinate domains (the latter form a subset of the former)!