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 (Michael). Under CH the Sorgenfrey line contains an uncountable subspace $X$ whose countable power $X^\omega$ is hereditarily Lindelof and hence 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 Sorgenfrey line the square $X\times X$ has uncountable spread.

Now I have a (relatively simple) ZFC-answer to the initial problem, see this preprint for more details.

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 $z(V_z\cap V_{f(z)})f(z)$ of the point $z\cdot f(z)$ in $G$. We claim that $x\cdot f(x)\notin z(V_z\cap V_{f(z)})f(z)$ for any $x\in Z\setminus\{z\}$. To derive a contradiction, assume that $x\cdot f(x)\in z(V_z\cap V_{f(z)})f(z)$ 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 (Michael). Under CH the Sorgenfrey line contains an uncountable subspace $X$ whose countable power $X^\omega$ is hereditarily Lindelof and hence 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 Sorgenfrey line the square $X\times X$ has uncountable spread.

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.

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 (Michael). Under CH the Sorgenfrey line contains an uncountable subspace $X$ whose countable power $X^\omega$ is hereditarily Lindelof and hence 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 Sorgenfrey line the square $X\times X$ has uncountable spread.