Is $[0,1]^2$ with the dictionary order topology separable? First countable? I have question about order square $I^2_0$ which is $[0,1]\times[0,1]$ with dictionary order 
this is what I know about it: compact Hausdorff, hence normal and regular as well. and also it is Lindelof. Therefore, it is not second countable, and metrizable (the  argument is not difficult) just by playing by theorems. However, I want to whether or not it is separable. I do it by way of contradiction. suppose it is separable we know that each open subspace of separable is separable. let $Y$ be a space of $X$ and $Y=(0,1)\times(0,1)$ it is open. For each $x\in{Y}$ there exist $U_x={x}\times(0,1)$. It makes the subspace $Y$ not separable.
my question is Is my argument about separable correct? 
 and also how about first countable? 
 A: Subspaces of separable spaces do not need to be separable in general: See Sorgenfrey plane (it is not compact, but locally compact, so we can take its one-point/Alexandrov compactification, which is then compact, Hausdorff, separable, but has a non-separable subspace).
As for separability, the sets
$$U_x=x\times(0,1)=\left\{w\in I_0^2:(x,0)<w<(x,1)\right\}$$
are all open, disjoint, and there are uncountably many of them. So if $D$ is any countable set, there will be some (in fact, lots of) $U_x$ which does not intersect $D$, so $D$ is not dense and $I_0^2$ is not separable.
As for first-countability: Each $U_x$ is isomorphic to $(0,1)$, which is first countable. So an element which is in some $U_x$, i.e., an element of the form $(x,y)$ with $y\neq0,1$, has a countable basis.
So let's consider an element in the "boundary": Say we take an element of the form $(x,1)$. A basic neighbourhood of $(x,1)$ has the form $(w,v)=\left\{z:w<z<v\right\}$, where $w<x<v$ (for $(1,1)$ we don't consider $v$, but the argumens are the same). So take such $w=(w_1,w_2)$ and $v=(v_1,v_2)$.
If $w_1<x$, then $w<(x,0)<(x,1)$. If $w_1=x$, then $w_2<1$, so $w<(x,1-1/n)<(x,1)$ for all $n$ large enough
Also, we necessarily have $v_1>x$, so $(x,1)<(x+1/n,0)<v$ for all $n$ large enough.
This shows that the intervals of the form $((x,1-1/n),(x+1/n,0))$ form a (countable) neighbourhood basis for $(x,1)$.
Similarly, points of the form $(x,0)$ also admit countable neighbourhood bases, so $I_0^2$ is first countable.
