Metrizable space under lexicographic order topology While doing some exercises about different metrics, I had a doubt about my comprehension of the matter.
Recall that the lexicographic order on $\mathbb{R}^2$ is defined as follows :
$$
 (a,b) < (c,d) \iff (a<c) \quad \mathrm{or} \quad (a=c \ \mathrm{and} \ b<d)
$$
We put the order topology on the real plane, so the open intervals of the form  $](a,b), (c,d)[$ look like a vertical open interval if $a=c$ or the union of a vertical line over $(a,b)$ with the band of all the points $(x,y)$ such as $a<x<c$ and the vertical line at $x=c$ up to $y=d$. (I don't know if it makes sense, it was just to get the picture in a sentence... but drawing an example on a sheet of paper should also work.)
If I did not make any mistakes, the real plane with such a topology is metrizable (is it true ?).
For the record, I defined $d$ a metric as :
$$
 d( (x,y), (x',y') ) = 
\begin{cases}
\infty \ \mathrm{if} x \neq x' \\
|y'-y| \ \mathrm{otherwise}
\end{cases}
$$
I'll skip the proof (but if $d$ is not a good candidate, please feel free to correct me on that point) that $d$ is a metric and defines the same topology that we had from the lexicographic order.
Then, the next question was about the existence of a metric on $[0,1]^2$ that gives the same topology as the one from the order defined above. I think I did the usual proof that consists of saying that the square is compact and therefore, if it was metrizable, it would separable which is not the case with the lexicographic order topology (the line passing through the middle $\{(s,1/2)|s\in[0,1]\}$ is discrete and uncountable).
First, is that statement correct ?
Second, what prevent us to use to induced metric on the square ? I couldn't think of a counter example when someone told me that restricting $d$ to the square was enough to get a metric space. But, as it would contradict the "proof" above, one of them is wrong (or something in broken in the math world, which I highly doubt and fear).

The unit square is not compact in this metric.  Your example of the line through the middle proves this.  You just picked the wrong part of the antecendent of (metrizable and compact) $\implies$ separable to deny.
