Consider the set $M:=\mathbb{R}_0 \cup \{A,B\}$ where $A\neq B \notin \mathbb{R}$.
Define $\phi_1: U_1:= \mathbb{R}_0 \cup \{A\} \to \mathbb{R}: x \mapsto \begin{cases} x \quad x \neq 0 \\ 0 \quad x = A\end{cases}$
Define $\phi_2: U_2:= \mathbb{R}_0 \cup \{B\} \to \mathbb{R}: x \mapsto \begin{cases} x \quad x \neq 0 \\ 0 \quad x = B\end{cases}$
Consider the atlas $\mathcal{A}:= \{(U_1, \phi_1), (U_2, \phi_2)\}$
and consider the topology
$$\mathcal{T}:= \{V \subseteq M\mid \phi_i(V \cap U_i) \mathrm{ \ open \ in \ \mathbb{R} \ for \ i =1,2}\}$$
My textbook says that we define a manifold $(M, \mathcal{A})$ as a space where the natural topology is Hausdorff because this example is pathological.
Questions:
(1) I think this topology is non-Hausdorff. Can we separate the points $A$ and $B$ in this topology? I think not, but did not manage to find a rigorous proof.
(2) Why is this a pathological example?
Let me first of all say something about the usage of "pathological". This is usually used for "this is something we can construct but is not what usually appears in practice". I don't think that there is any precise definition of "pathological".
Now to your first question: I assume that $\mathbb R_0=\mathbb R\setminus\{0\}$. You are right, this space is not Hausdorff. To have a picture in mind, this is the real line with two origins. A base for your topology is provided by all open sets not containing the origin together with open sets of the form $(-\epsilon_1,0)\cup A\cup (0,\epsilon_2)$ with $\epsilon_1,\epsilon_2>0$ and $(-\epsilon_1,0)\cup B\cup (0,\epsilon_2)$ with $\epsilon_1,\epsilon_2>0$. Every neighborhood of $A$ (respect. $B$), say $I=(-\epsilon_1,0)\cup A\cup (0,\epsilon_2)$ is naturally homeomorphic to the open interval $(-\epsilon_1,\epsilon_2)$. Now whatever neighborhood of $A$ or $B$ you take, you can map it to such an interval via an homeomorphism and these intervals always intersect. So your space cannot be Hausdorff.
Are you fine with this explanation?