Is $d(x,y) = \sqrt{|x-y|}$ a metric on R? For $x,y \in \mathbb{R}$, define $d(x,y) = \sqrt{|x-y|}$.
Is this a metric on $\mathbb{R}$?
It's clear that $d(x,x) = 0$ and $d(x,y) = d(y,x)$ for all $x,y \in \mathbb{R}$.  The triangle inequality seems to hold for all values I have tested, but I have not found this function anywhere online as an example of a metric on $\mathbb{R}$.

Sure looks like it. It's translation invariant, so to prove the TE for $x \le y \le z$, adjust everything so that the lowest, $x$, of the three values is at $0$ (i.e., add $-x$ to all three numbers). Then you need to show that 
$$
\sqrt{y} + \sqrt{z} \ge \sqrt{y+z}
$$
for any nonnegative $y$ and $z$, which is true (by just squaring both sides). 
