Let $$A=\{(x,y)\in\mathbb{R}^2 \ : \ |x|\leq 2 \}$$ $$B=\{(x,y)\in\mathbb{R}^2 \ : \ 0< x+y < 2 \}$$
and
$$C=A\cap B$$
One can define the following metric $d:\mathbb{R}^2\to \mathbb{R}$, by the rule $$d((x_1,x_2),(y_1,y_2))=|x_1-y_1|+d_0(x_2,y_2)$$ where $d_0$ stands for the discrete metric defined as $d_0(x, y) = 0$ if $x = y$ and $d_0(x, y) = 1$ otherwise.
The question asks for describing the closure of set $C$ under metric $d$ defined above.
Here is my attempt understanding this problem:
First question that I asked myself is 'how I can simplify $d$?'
One can write $d((x_1,x_2),(y_1,y_2))=|x_1-y_1|$ if $x_2=y_2$, and $d((x_1,x_2),(y_1,y_2))=|x_1-y_1|+1$ otherwise. Next, I asked 'how can I describe an closed ball of such metric?' So, if $(q_1,q_2)\in\mathbb{R}^2$ and $r=0$, then $B_d[(q_1,q_2),r]=(q_1,q_2)$. Next, suppose that $r>0$. And then, $$B_d[(q_1,q_2),r]=\{(x_1,x_2)\in\mathbb{R}^2:d((q_1,q_2),(x_1,x_2))\leq r\}$$ which is $$\{(x_1,x_2)\in\mathbb{R}^2:|x_1-q_1|+d_0(x_2,q_2)\leq r\}$$, so, if $x_2=q_2$, then $d_0(q_2,x_2)=0$ and we have $$\{(x_1,x_2)\in\mathbb{R}^2:|x_1-q_1|\leq r\}=[q_1-r,q_1+r]\times\{q_2\}.$$
But if $x_2\neq q_2$, then $d_0(q_2,x_2)=1$ and
$$\{(x_1,x_2)\in\mathbb{R}^2: x_2\neq q_2 \ and \ |x_1-q_1|+d_0(x_2,q_2)\leq r\}$$,
which is
$$\{(x_1,x_2)\in\mathbb{R}^2: x_2\neq q_2 \ and \ |x_1-q_1|\leq r-1\}$$,
so
if $r<1$ we have $B_d[(q_1,q_2),r]=\emptyset$, while if $r\geq 1$ then
$$[q_1-(r-1),q_2+(r-1)]\times (\mathbb{R}-\{q_2\}$$
How I can proceed from here, or perhaps there is a different way, describing the closure of $C$ under $d$?