I've to prove the following result but I'm not sure about my solution.
Prove that the topology on a manifold $M$ is of Hausdorff iff it admits an atlas $\mathcal{A}=\{U_{\alpha}, \phi_{\alpha}\}$ with the following property: $\forall$ couple of points $p,q\in M$ there exists a chart such that $p,q \in U_{\alpha}$ or two charts such that $p\in U_{\alpha}$ and $q\in U_{\epsilon}$ and $U_{\alpha}\cap U_{\epsilon}=\emptyset$.
Proof Let we suppose to have a Hausdorff topology on $M$. Then we have that for each couple of distinct points $p, q \in M$ there exist two open sets, $A$ containing $p$ and $B$ containing $q$, such that $A \cap B =\emptyset$. The topology on $M$ is induced by the charts i.e. $U_{\alpha}\in \mathcal{A}$ is open iff $\phi_{\alpha}(U_{\alpha})$ is open in $\mathbb{R}^{n}$. Let we consider two scenario. If $\phi_{\alpha}(U_{\alpha})$ is open in $\mathbb{R}^{n}$ and contains both $\phi_{\alpha}(p)$ and $\phi_{\alpha}(q)$ such that $\phi_{\alpha}(q)\neq \phi_{\alpha}(p)$ we are able two find two other charts $U_{\beta}$ and $U_{\gamma}$ in $\mathcal{A}$ such that $\phi_{\alpha}(q) \in \phi_{\alpha}(U_{\alpha}\cap U_{\beta})$ $\phi_{\alpha}(p)\in \phi_{\alpha}(U_{\alpha}\cap U_{\gamma})$ and $\phi_{\alpha}(U_{\alpha}\cap U_{\gamma})\cap \phi_{\alpha}(U_{\alpha}\cap U_{\beta})= \emptyset$. If would not be the case, $M$ could not has the Hausdorff property, which needs to be true for each chart in the atlas. Since $\phi_{\alpha}$ are homomorphisms, the same holds for $U_{\gamma}, U_{\alpha}$ and $U_{\beta}$ i.e. $(U_{\alpha}\cap U_{\gamma})\cap (U_{\alpha}\cap U_{\beta})= \emptyset$.
The same reasoning holds for $\phi_{\alpha}(q)\neq \phi_{\epsilon}(p)$ belonging to different open sets of $\mathcal{A}$, $U_{\alpha}$, $U_{\epsilon}$.
Suppose the by contradiction that $M$ has not the Hausdorff property. This means that we cannot find out at least two open set in the atlas, containing two distinct points, such that their intersection is void. Since holds $\phi_{\alpha}(U_{\alpha}\cap U_{\gamma})\cap \phi_{\alpha}(U_{\alpha}\cap U_{\beta})= \emptyset$ for $\phi_{\alpha}(q)\neq \phi_{\alpha}(p)$, $\phi_{\alpha}(q)\in \phi_{\alpha}(U_{\alpha}\cap U_{\beta}), \phi_{\alpha}(p) \in \phi_{\alpha}(U_{\alpha}\cap U_{\gamma})$ from the previous case and $\phi_{\alpha}(U_{\alpha}\cap U_{\gamma})$ and $\phi_{\alpha}(U_{\alpha}\cap U_{\beta})$ are open as theri inverse image (which exists and it's open sine $\phi_{\alpha}$ is an homomorphism), we get an contradiction. The same holds for the second case.
Let me know if the proof is correct. Thanks.