Check axioms of separation $T_{0},\, T_{1},\, T_{2}, \,T_{3}, \, T_{3\frac{1}{2}}\, T_{4}$ for $(x - r, x + r)\backslash \{y = x \pm \frac{1}{n}: n \in \mathbb{N}\},\, x \in \mathbb{R},\, r>0 $.
My work:
$T_{0}$: these sets satisfied;
$T_{1}$: these sets satisfied;
$T_{2}$: $x \in U, y \in V, \quad U \cap V = \emptyset$
Let $x \in (x - \varepsilon, x + \varepsilon), \, y \in (y - \varepsilon, y + \varepsilon)$. $U \cap V = \emptyset?$
$(x - \varepsilon, x + \varepsilon) \cap (y - \varepsilon, y + \varepsilon) = \emptyset$. So, $T_{2}$ aslo satisfied.
$T_{3}$:
Let $x \in (x - \varepsilon, x + \varepsilon) = U$
$V = (x - \frac{\varepsilon}{2}, x + \frac{\varepsilon}{2})$
$x \in (x - \frac{\varepsilon}{2}, x + \frac{\varepsilon}{2}) \rightarrow x \in V \subset \overline{V} \subset U$. So, also satisfied.
$T_{3\frac{1}{2}}$ and $T_{4}$ - here I dont know how to check it
You presumably mean the topology that has as a local base at $x$ all sets of the form $(x-\varepsilon, x+\varepsilon)\setminus \{x\pm \frac{1}{n}: n \in \mathbb{N}\}$, where $\varepsilon >0$.
Check that $F=\{\frac1n, n=1,2,3,\ldots\}$ is closed in this topology but we cannot separate it from $0 \notin F$. So it is not $T_3$, $T_{3\frac12}$, nor $T_4$. It's comparable to Munkres' $\mathbb{R}_K$ (page 200, example 1, first edition).