Now I have a (relatively simple) ZFC-answer to the initial problem.

>**Theorem 1.** If a topological group $G$ contains a topological copy of the Sorgenfrey line, then it contains a discrete subspace of cardinality continuum.

*Proof.* Let $X$ be a subspace of $G$, homeomorphic to the Sorgenfrey line. Then $X$ contains a subset $Z\subset X$ of cardinality $|Z|=\mathfrak c$ and a function $f:Z\to X$ which is strictly decreasing with respect to a linear order $\le$ on $X$ such that for every $x\in X$ the set ${\uparrow}x:=\{y\in X:x\le y\}$ is a closed-and-open neighborhood of $x$ in $X$. Without loss of generality, the set $Z$ has no maximal element. 
For any point $x\in X$ choose a neighborhood $V_x\subset G$ of the unit $e$ of $G$ such that $X\cap (V_x^{-1}V_xx\cup xV_xV_x^{-1})\subset {\uparrow}x$.

The space $X$, being homeomorphic to the Sorgenfrey line is (hereditarily) separable and hence contains a countable dense subset $C$.

Then for every point $x\in X$ we can choose a point $u_x\in C\cap{\uparrow}x$ such that 
the order interval $[x,u_x):=\{y\in X:x\le y< u_x\}$ is a neighborhood of $x$ in $X$ such that $[x,u_x)\subset xV_{f(x)}$ if $x\in Z$ and $[x,u_x)\subset V_{f^{-1}(x)}x$ if $x\in f(Z)$. Since the continuum $\mathfrak c$ has uncountable cofinality, for some $c,d\in C$ the set $\{z\in Z:u_z=c,\; u_{f(z)}=d\}$ has cardinality $\mathfrak c$. Replacing $Z$ by this uncountable set, we can assume that $u_z=c$ and $u_{f(z)}=d$ for all $z\in Z$.

We claim that the subspace $D:=\{z\cdot f(z):z\in Z\}\subset G$ has cardinality $\mathfrak c$ and is discrete in $G$. For every $z\in Z$ consider the neighborhood $zV_{f(z)}f(z)$ of the point $z\cdot f(z)$ in $G$. We claim that  $x\cdot f(x)\notin zV_{f(z)}f(z)$ for any $x\in Z\setminus\{z\}$. To derive a contradiction, assume that $x\cdot f(x)\in zV_{f(z)}f(z)$ for some $x\ne z$ in $Z$.

If $x>z$, then $x\in [z,u_z)\subset zV_{f(z)}$ and $$f(x)\in x^{-1}xf(x)\in x^{-1}zV_{f(z)}f(z)\subset V_{f(z)}^{-1}z^{-1}zV_{f(z)}f(z)=V_{f(z)}^{-1}V_{f(z)}f(z).$$Then $f(x)\in X\cap V_{f(z)}^{-1}V_{f(z)}f(z)\subset {\uparrow}f(z)$ and $f(x)\ge f(z)$, which is not possible as $x>z$ and $f$ is strictly decreasing.

If $z>x$, then $f(x)>f(z)$ and $f(x)\in [f(x),u_{f(x)})=[f(x),d)\subset [f(z),d)= [f(z),u_{f(z)})\subset V_{z}f(z)$ and then
$$x\in zV_zf(z)f(x)^{-1}\subset zV_zf(z)f(z)^{-1}V_z^{-1}=zV_zV_z^{-1}\subset{\uparrow}z,$$ which contradicts $z>x$. $\square$

By analogy we can prove the following more refined version of the above theorem.

>**Theorem 2.** If a topological group $G$ contains a subspace $X$, homeomorphic to a subspace of the Sorgenfrey line, then $s(G)\ge s(X\times X)$.

Here $s(X)=\sup\{|D|:D$ is a discrete subspace in $X\}$ is the *spread* of a topological space $X$. 

To my big surprise I have discovered that for an uncountable subspace $X$ of the Sorgenfrey line the spread $s(X\times X)$ is not necessarily uncountable.

>**Theorem 3.** Under CH the Sorgenfrey line contains an uncountable subspace $X$ whose square $X\times X$ has countable spread.

On the other hand, Theorem 4.8(c) of Todorcevic's book "Partition Problems in Topology" implies an opposite 

>**Theorem 4.** Under OCA for any uncountable subspace $X$ of the real line the square $X\times X$ has uncountable spread.