One can also ask for a weaker form of metrizability, where distances are measured in any ordered field (not neccesarily the reals). A example for such a bigger ordered field is the hyperreals http://en.wikipedia.org/wiki/Hyperreal_number. Lets call them $\mathbb{R}^*$.

For example the hyperreals are not $\mathbb{R}$-metrizable, but they are $\mathbb{R}^*$-metrizable just by setting $d(a,b):=|a-b|$. So we have already a example for a nonmetrizable space.

Given any ordered field $F$ and a $F$-metrizable space $S$. Then one can find a local basis at every point $s\in S$ via $\{B_\varepsilon(s)| \varepsilon \in F,\varepsilon >0\}$. Note that this basis is totally ordered (under the inclusion).

Now consider the space $X:=\prod_{i\in I} \{0;1\}$, where $I$ is a uncountable set and let $x\in X$ be any point. One can show, that there is no local basis at $x$, that is totally ordered. So this space is not metrizable for any ordered field $F$.

I think this is quite surprising. Heuristically speaking, this space is too big to be $\mathbb{R}$-metrizable. But it is even not $F$ - metrizable for bigger $F$'s.

It shows, that Urysohns metrization theorem cannot be generalized to $\mathbb{R}^*$-metrics, i.e. a statement like

"A regular Hausdorff space with a basis for the topology of cardinality $\le$ ? is $ \mathbb{R}^*$-metrizable. "