Sample of topology distinguishable but not separated. I'm confused with the definition of $R_0$ space that any two topologically distinguishable points are separated.
Is there a sample of two topologically distinguishable points are not separated ?
 A: The simplest example of a space like that is a Sierpiński space, say $X=\{0,1\}$ with topology 
$\{\emptyset, \{0\},X\}$. The two points $0$ and $1$ are topologically distinguishable: their sets of neighbourhoods are different: $\{0\}$ has neighbourhoods $\{0\}$ and $X$, and $1$ has only $X$. 
But they cannot be separated as $1$ does not have an open neighbourhood that does not contain $0$.
Recall that $x\neq y \in (X, \mathcal{T})$ are topologically distinguishable iff
$$\left(\exists U \in \mathcal{T}: (x \in U \land y \notin U)\right) \lor \left(  \exists U \in \mathcal{T}: (x \notin U \land y \in U) \right)\tag{1}$$
which is a formal way of saying that $\mathcal{O}(x) \neq \mathcal{O}(y)$, where those symbols denote the set of open neighbourhoods of $x$ resp. $y$.
And $x \neq y$ are called separated iff
$$\left(\exists U \in \mathcal{T}: (x \in U \land y \notin U)\right) \land \left(  \exists U \in \mathcal{T}: (x \notin U \land y \in U) \right)\tag{2}$$
So it's clear that (instead of "and" we have "or"):

"$x$ and $y$ are separated" $\implies$ "$x$ and $y$ are topologically distinguishable" $\implies$ $x \neq y$.

And $X$ is $T_0$ iff we can reverse the last arrow, and $X$ is $R_0$ iff we can reverse the first arrow. 
And a space is $T_1$ iff the first and last statement are equivalent, and so $R_0 + T_0 \iff T_1$. So we will have examples of such points as you aked for in any space that is $T_0$ but not $T_1$, like Sierpi&nacute;ski space. 
Slightly more interesting examples: $X=\mathbb{N}=\{0,1,2,\ldots\}$ in the upper topology $\{\emptyset, n^\uparrow (n \in \mathbb{N})\}$, where $n^\uparrow=\{m \in \mathbb{N}: m \ge n\}$ and the same $X$ in the lower topology $\{X, \emptyset, n^\downarrow (n \in \Bbb N)\}$, where $n^\downarrow=\{m \in \Bbb N: m \le n\}$. In both spaces, all pairs of distinct points are topologically distinguishable, but no such pair is separated. Fun fact: any infinite $T_0$ space that is not $T_1$ will contain one of these as a subspace, BTW. 
As an extreme (non-)example: let $X$ $(|X| \ge 2$) be any set in the indiscrete (trivial) topology: in this $X$ there are no points that are topologically indistinguishable (all have $\mathcal{O}(x)=X$) so voidly $X$ is $R_0$ (all pairs of indistinguishable points can be separated because there are no such pairs).
The example by GreginGre (the double-pointed countably infinite cofinite space) is an example of a different kind: it shows $R_0$ (which this space is) does not imply either $T_0$ nor $T_1$, so the R-axioms (there is also $R_1$) are "independent' of the T-axioms, so to say.
A: The double point cofinite topology seems to be the example you are looking for. 
See 
https://en.wikipedia.org/wiki/T1_space
and
https://en.wikipedia.org/wiki/Cofiniteness#Double-pointed_cofinite_topology
