About a metric space

Question

If we define a metric $d(x,y)=\begin{cases}|x|+|y|,&\text{if }x\neq y\\0,&\text{if }x=y\end{cases}$ on $\mathbb{R}$, which of following options are true?

1. Every compact set is finite,
2. $\mathbb{N}$ is open,
3. $(\mathbb{R},d)$ is connected,
4. $\mathbb{Z}$ is open.

I think we have $ B_r(n)=\{ n\}, n\in\mathbb{N}, r <n$ and $ B_r(0)=(-r,r)$ (note that $ B_r(x_0)= \{y\in\mathbb{R} | d(y,x_0)<r \}$) so $ \mathbb{N}$ is open and $ \mathbb{Z}$ is not open.

$(\mathbb{R},d)$ is not connected because $ \mathbb{N}$ is open and close.

Note that $ k(x,y)=\lvert x+y\rvert \le \lvert x\rvert+\lvert y\rvert =d(x,y) $ then every open set in $(\mathbb{R},k)$ is open in $(\mathbb{R},d)$.

Now is a compact set in $(\mathbb{R},k)$ also compact in $(\mathbb{R},d)$ ?

Answer 1

1. The set$$C=\{0\}\cup\left\{\frac1n\,\middle|\,n\in\mathbb N\right\}$$is compact, because any ball centered at $0$ contains all but finitely many elements of $C$. Therefore, the first assertion is false.
2. True, since the distance between a natural number $n$ and any real number (other than $n$) is greater than $1$. Therefore,$$\mathbb{N}=\bigcup_{n\in\mathbb{N}}B_1(n).$$
3. The set $\{1\}$ is both closed and open. Since, furthermore, $\emptyset,\mathbb{R}\neq\{1\}$, $(\mathbb{R},d)$ is not connected.
4. This is false. Every open ball centered at $0$ contains real numbers distinct from $0$.