It is easy to prove that the completion of your space can be any separable metric space
where metric spheres are nowhere dense.

Does not it scare you?

**Answer to the comment:**
Not all of your spaces can be embedded into a metric tree.

BTW, there is a nice characterization of subsets in a metric tree:
$$ | x - y | + | a - b | \le \max\{|x-a| + |y-b|,|x-b|+|y-a|\}$$
for all points $x,y,a,b$
(here $|x-y|$ denotes the distance from $x$ to $y$).

In other words, the values
$$X=|x-a|+|y-b|,$$
$$Y=|x-b|+|y-a|,$$
$$Z=|x-y|+|a-b|$$ satisfy ulrtatriangle inequality.
This inequality can be also thought as a discrete analog of CAT[−∞] inequality.