Proving Sorgenfrey line not homeomorfpic to $\mathbb{R}$ and $\mathbb{R}^2$ and its subspaces 
Let $\mathscr{B}=\{[a,b):a\in\mathbb{R}, b\in\mathbb{Q},a<b\}$ define a topology $\tau_1$. Prove that the Sorgenfrey line($(\mathbb{R},\tau_1)$) is not homeorphic to $\mathbb{R},\mathbb{R}^2$
  ,or any subspace of either spaces.

For me this claim seems counter-intuitive once the Sorgenfrey line is the real line. 
I try to use connectedness but all the spaces aforementioned are connected. I have no clue on how to solve the problem.
Question:
Can someone provide me hints?   
Thanks in advance! 
 A: The Sorgenfrey line is separable ($\mathbb{Q}$ is dense in it) but not second countable: if $\mathcal{B}$ is any base for the Sorgenfrey topology, then pick $B(x) \in \mathcal{B}$ such that $x \in B_x \subseteq [x,x+1)$. This can be done by virtue of $\mathcal{B}$ being a base for that topology, and the fact all sets $[x,x+1)$ are open.
If then $x < y$, then $x \notin B(y)$ while $x \in B(x)$ so $B(x) \neq B(y)$. This shows that all $B(x)$ are distinct for distinct $x \in \mathbb{R}$. So every base for the Sorgenfrey line is uncountable (even of size $\mathfrak{c} = |\mathbb{R}|$).
In a metrisable space $X$, if $X$ is separable it has a countable base.
So the Sorgenfrey line cannot be homeomorphic to (any subspace of) a metrisable space, including $\mathbb{R}$ or the plane in its usual topologies.
A: Hint: the Sorgenfrey line is not connected, in fact it is totally disconnected, namely any subspace with more than one element fails to be connected. 
As pointed out by Henno though some subspaces or $\mathbb{R}$ are totally disconnected, so this won't give you the full answer you are looking for. 
Different approach: the Sorgenfrey line is not second countable.
