In the Sorgenfrey plane, is an open disc homeomorphic to an open square? In the sorgenfrey plane $\mathbb{R}_l^2$, the subspace
$$X=\{(x,y):x^2+y^2\leq 1\}$$
is not homeomorphic to the subspace
$$Y=\{(x,y):|x|\leq 1,|y|\leq 1\},$$
because there is only one isolated point in $Y$ but infinitely many in $X$.
Now we modify the question a little bit. Is the subspace
$$X=\{(x,y):x^2+y^2<1\}$$
homeomorphic to the subspace
$$Y=\{(x,y):|x|<1,|y|<1\}?$$
 A: Edit: Here is a general theorem that does the job:
Theorem. Let $U, U'$ be two nonempty subsets of $R^n$ which are open in the standard topology. Then they are homeomorphic with respect to the Sorgenfrey topology.
Proof. First, a notation: I will call a basic open set in Sorgenfrey topology, i.e. a product of right-open intervals, an S-box. 
A product map of an S-box $B$ in $R^n$ is a map of the form
$$
F(x_1,...,x_n)=(f_1(x_1),...,f_n(x_n)).
$$
Suppose that each component $f_i$ of such a map is the restriction of a strictly increasing map of the real line, continuous in the standard topology. Then $f_i$ 
is also continuous in Sorgenfrey topology. Hence, $F: B\to F(B)$ is a homeomorphism in Sorgenfrey topology. I will use this observation in the case of affine product maps $F$.  
Here is the key geometric lemma that I will need:
Lemma. Let $U$ be a nonempty subset of $R^n$ which is open set in the standard topology. Then $U$ can be represented as a countably-infinite disjoint union of S-boxes. 
Proof. Start with a maximal collection $C_1$ of pairwise disjoint S-boxes contained in $U$, whose vertices have integer coordinates. Next, take a maximal collection $C_2$ of disjoint S-boxes in $U$, so that the new boxes are disjoint from the boxes in $C_1$ and have vertices with coordinates in 
$\frac{1}{2}{\mathbb Z}$. Repeat this using S-boxes with vertices in 
$\frac{1}{4}{\mathbb Z}^n$. Continue inductively. qed   
I now continue with the proof of the theorem. I will denote the collection of boxes constructed in this lemma by $C(U)$. Suppose now that we have two nonempty sets $U, U'$ of $R^n$ which are open in the standard topology. Define the coverings $C(U)$, $C(U')$ for these sets: 
$$
C(U)=\{B_i: i \in {\mathbb N}\}, C(U')=\{B_i': i \in {\mathbb N}\}. 
$$
Then for each $i$ take a product affine homeomorphism $F_i: B_i \to B_i'$. 
This defines a bijection $F: U \to U'$. Each $F_i$ is a homeomorphism 
 in Sorgenfrey topology and each set $B_i, B_i'$ is open in Sorgenfrey topology. 
Therefore, $F: U \to U'$ is a homeomorphism in Sorgenfrey topology. qed 
On the other hand, the following is also true:
Theorem. Let $U, U'$ be nonempty subsets of $R^n$ which are open and connected in the standard topology. Suppose that $F: U \to U'$ is a homeomorphism in Sorgenfrey topology as well as in the standard topology. Then $F$ is a product map. 
Thus, there is no map between open square and an open round ball which is a homeomorphism in Sorgenfrey topology as well as in the standard topology.
My first attempted proof below was wrong, I keep it so that the comments underneath my answer make sense.
Yes, they are homeomorphic. More generally, let $B^n(R)$ denote the open (in standard topology) $R$-ball in $R^n$ and $\bar{B}^n(R)$ the closed (in standard topology) $R$-ball in $R^n$. Then 
Theorem. $B^n(1)\times B^m(1)$ is homeomorphic to $B^{n+m}(1)$ with respect to the product Sorgenfrey topology on $R^{n+m}$. 
Proof. From now on, everything is done with respect to the product Sorgenfrey topology on suitable $R^k$. 
First note that $B^n(R)$ is open and $\bar{B}^n(R)$ is closed. Furthermore,  every dilation of $R^k$ is a homeomorphism. 
Now, I claim that $B^n=B^n(1)$ is homeomorphic to $R^n$. Indeed, exhaust $B^n$ by the balls $B^n(1-1/i)$ and exhaust $R^n$ by the balls $B^n(i)$. Now, glue dilations of spherical shells to construct a homeomorphism $B^n\to R^n$. (Here we use the gluing lemma for continuous maps: If $f: X=X_1\cup X_2\to Y$ is continuous on closed subsets $X_1, X_2$ then it is continuous on $X$.) 
Lastly, use product of the above homeomorphisms to show that $B^n\times B^m$ is homeomorphic to $R^{n+m}$. Hence,
$$
B^{n+m}\cong R^{n+m}\cong B^n\times B^m. 
$$
qed
