Metrizable spaces and the order topology I am trying to find a solution for the following exercise.

Let $(X,\leq)$ be a total (=linear) order.
a) Let B=$\{(a,b): a,b \in X \ \cup \{-\infty, \infty \} \}$. There is a topology T on X with base B.
b) Is (X,T) metrizable, i.e can be a metric d on X be given such that the induced topology coincides with T?

My problem lies with b). I know that  a compact Hausdorff space is metrizable if and only if it is second-countable.
The unit square equipped with the lexicographical order is compact, but not second-countable and hence cannot be metrizable, proving that the answer to the question is not always "yes". My question is: Is it possible to give an example that does not rely on second - countability or separability? Can I, say, construct a space with an order topology such that the space is not first - countable (and hence cannot have a metric)?
 A: Part a) is actually quite sloppy. The standard base for a linearly ordered space (which need not, but can have a maximum (say $+\infty$ if there is one) or a minimum (if there is, I'll call it $-\infty$) is all sets of three possible types:
$$\{(a,b): a,b \in X\}, \{[-\infty, a): a \in X\}, \text{ if } -\infty \in X, \{(a,+\infty], a \in X\}, \text{ if } +\infty \in X$$
So in a set like $X=\mathbb{R}$ we only need the first type, while in $X=\mathbb{N}$ we also have the second type (as $0$ functions as $-\infty$, the minimum of $X$). The lexicographically ordered square $(X=[0,1]^2, \le_{\mathrm lex})$ has all three types with $-\infty = \min(X)=(0,0)$ and $+\infty= \max(X)=(1,1)$, etc. 
It is actually a classic well-studied problem when ordered spaces have a compatible metric, and indeed one of the most accessible ones is the lexicographically ordered square (which you mention) which is compact but not separable (while any compact and metrisable space is always separable).
If it's a non-first countable ordered space you like the easiest one is the successor ordinal of the first uncountable ordinal, which Munkres denotes by $\overline{S}_\Omega$ but in more common notation is denoted $\omega_1 + 1$. The maximal element of this space (i.e. $\omega_1$) does not have a countable local base (but all other points do; many are even isolated points).
An ordered space is metrisable iff it has a $G_\delta$ diagonal ($\Delta_X=\{(x,x): x \in X\}$ is a countable intersection of open sets of $X^2$ or iff it has a $\sigma$-locally countable base. See the encyclopaedia of General Topology for references, e.g.  
