It's easy to check if a graph can be embedded in the plane: just check for forbidden minors. Is it also easy to check if a "distance function" can be embedded? Are there any necessary and sufficient conditions one can check?
I know there's a lot of research into approximate embeddings, but clearly in some cases that's not necessary.
There are very high obstacles to isometric embedding a metric space in the plane! If there exist three paints $a,b,c$ where the triangle inequality is strict (i.e. $d(a,b)+d(b,c)+d(a,c)<2\max\{d(a,b),d(b,c),d(a,c)\}$) then any other point is "very likely" to produce a contradiction: From knowing $d(a,x)$ and $d(b,x)$ there are only two candidate positions for $x$ and it really needs a "coincidence" if one of them has the correct distance from $c$.
I have the following
Conjecture. A metric space $(X,d)$ can be isometrically embedded in the plane $\mathbb R^2$ endowed with the standard metric $\rho$ iff each four-point subspace of $(X,d)$ can be isometrically embedded in the plane.
To build such an embedding $i$ we fix three different points $x$, $y$ and $z$ of $X$ for which the triangle inequality is strict and fix an arbitrary embedding $i_0:(\{x,y,z\},d|\{x,y,z\})\to (\mathbb R^2, \rho )$ (case when there are no such triple $\{x,y,z\}$ have to be considered separately). Let $t\in X$ be an arbitrary point. It seems that there exists an unique point $t’\in \mathbb R^2$ such that $d(x,t)=\rho(i_0(x),t’)$, $d(y,t)=\rho(i_0(y),t’)$, and $d(z,t)=\rho(i_0(z),t’)$. Put $i(t)=t’$. Then it seems that $i$ is an isometric embedding.