Let $R$ be a set of points and $\mathbb{D}$ be a totally ordered field. Further consider a function $\rho:R\times R \rightarrow \mathbb{D}$.
$\langle R,\mathbb D,\rho\rangle$ is a metric space if $\rho$ has the following properties:
1) $\rho(x,y) \geqslant 0$
2) $\rho(x,y) = 0$ iff $x = y$
3) $\rho(x,y) = \rho(y,x)$
4) $\rho(x,z) \leqslant \rho(x,y) + \rho(y, z)$
Does this less restricted definition properly induce a topology on $R$ in the way expected by a standard metric and are there spaces which metrizable under this concept of metric that are not metrizable under the usual restriction?
In general, these two concepts of metrisability are different. Consider an ultrapower $\mathbb{D}$ of $\mathbb{R}$ over a non-principal ultrafilter on $\mathbb{N}$. It is readily seen that $\mathbb{D}$ is non-archimedean: consider the element
$$[(1,2,3,4, \ldots )].$$
I claim that each basis of neighbourhoods of the origin induced from your non-standard metric is uncountable. If there were a countable basis of the origin, $\mathbb{D}$ would be archimedean. This proves that $\mathbb{D}$ with the corresponding topology is not first countable, hence non-metrisable.