Unusual Topology, interior and closure 
On $\mathbb{R}^2$ let $\tau$ the collection of subsets defined as it follows $$A \in \tau \iff (x,y) \in A \implies (y,x) \in A$$
let $$B = \{(x,y) \in \mathbb{R}^2 : y = x\}, C = \{(x,y) \in \mathbb{R}^2 : y = -x\}$$
$D = \{(x,y)\in B\cup C : x \in [-2,3)\}$, find interior, closure and frontier of $D$.

I've noticed that I struggle very much in those kind of exercises, I'm wondering if exists a better approach.
My attempt: After verifying that $\tau$ is actually a topology on $\mathbb{R}^2$, I was thinking that the biggest open set contained in $D$ is the square in $\mathbb{R}^2$, $Q = \{(x,y) \in \mathbb{R}^2 : -2 ≤ x≤ 2, -2 ≤ y≤ 2\}$. For the closure I'm not completely sure but I was thinking of $D$ itself and so the frontier should be:
$$\{(x,y) \in B\cup C: x \in (2,3)\}$$
Am I wrong? Thank you all very much
 A: First notice that the set $\{x\in B:x\in[-2,3)\}$ is open . And $\{x\in C: x\in [-2,2]\}$ is open. But if we take a point $(x_{0},y_{0})\in C$ such that $2<x_{0}<3$ . Note that any open set containing $(x_{0},-x_{0})$ must contain $(-x_{0},x_{0})$ and hence cannot be contained in $D$ and hence these points are not interior points.An open set containing such $(x_{0},-x_{0})$  is $\{(x_{0},-x_{0}),(-x_{0},x_{0})\}$ which does not intersect $D\setminus\{(x_{0},-x_{0})\}$ and hence these points are isolated points.
Note that for any point not in $D$ say $(x,y)$ we have the set $\{(y,x),(x,y)\}$ is an open set such that this set is disjoint from $D$ . Hence they cannot be part of the closure of $D$.
Thus $\text{int}(D)=\{x\in B:x\in[-2,3)\}\cup\{x\in C:x\in[-2,2]\}$ .
We have shown that any point outside $D$ cannot be a limit point and hence $\text{cl}(D)=D$ .
The frontier then just consists of the isolated points which are $\{x\in C:x\in (2,3)\}$ .
Note that this also matches with the formula that $\partial(D)=\text{cl}(D)\setminus\text{int}(D)$ . So if you manage to calculate interior and closure you can get the frontier for free.
