A metric such that $d(0,1)>1000d(0,2)$

I need to find a metric on $[0,2]$ such that $d(0,1)>1000d(0,2)$

Here is the the example I came up with the following $d(x,y)=\begin{cases} 0\ \ x=y \\ 1 \ \ x\neq y+1\\ 1001 \ \ x=y+1 \end{cases}$ I'm just not sure about the following property $d(x,y) \le d(x,p)+d(p,y)$ and what if $x>1$ . Will this metric work? If not can someone point out to a metric that will work. Thank you

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It will not work, as $d(0,0.5)+d(0.5,1) = 2 < d(0,1) = 1001.$
Let $f(x)$ be defined as $x$ for $x \in [0,1)$ and $\frac{2}{x-0.999}$ for $x \in [1,2]$. Define $d(x,y)$ as $|f(x)-f(y)|.$ Then, $d(0,1) = 2000$ and $1 < d(0,2) < 2.$
This metric satisfies the triangle inequality as $|f(x)-f(y)| + |f(y)-f(z)| \ge |f(x)-f(z)|,$ because absolute value satisfies the triangle inequality. If $d(x,y) = 0$, then $f(x) = f(y)$, but $f$ is in $[0,1]$ if $x\in [0,1)$ and is in $(1,2000]$ if $x\in [1,2].$ And $d(x,y) > 0$ for all $x,y$, as $d(x,y)$ is an absolute value.