In Orderability in the presence of local compactness, Valentin Gutev states and proves the following proposition:
A suborderable space $X$ is orderable with respect to a linear order $\prec$ on it if and only if each clopen $(\prec )$-cut of $X$ is either a gap or a jump.
In his proof, he shows that $\mathscr{T}$ the topology on $X$, and $\mathscr{T}_{\prec}$ coincide, i.e. $\mathscr{T} = \mathscr{T}_{\prec}$. In order to show that, he takes a point $x\in X$ such that $[x,\rightarrow)\in\mathscr{T}$ and concludes $[x,\rightarrow)\in\mathscr{T}_{\prec}$, hence $\mathscr{T} = \mathscr{T}_{\prec}$.
My question is, why is this enough, why does he not take an arbitrary open set in $X$?
Because $\langle X,\mathscr{T}\rangle$ is suborderable by $\preceq$, the topology $\mathscr{T}$ has a base $\mathscr{B}$ of $\preceq$-intervals. These intervals can be of any form: $(x,y)$, $(x,\to)$, $(\leftarrow,x)$, $(x,y]$, $[x,y)$, $[x,y]$, $[x,\to)$, or $(\leftarrow,x]$. To show that $\mathscr{T}=\mathscr{T}_{\preceq}$, it suffices to show that $\mathscr{B}\subseteq\mathscr{T}_\preceq$. This is certain true of all intervals of the first three types, so we need only show that any interval of one of the last five types that is in $\mathscr{T}$ is also in $\mathscr{T}_\preceq$.
First observe that essentially the same argument used in Gutev’s proof of Lemma $\mathbf{6.4}$ shows that if $x\in X$ is such that $(\leftarrow,x]\in\mathscr{T}$, then $(\leftarrow,x]\in\mathscr{T}_{\preceq}$. Thus, any interval of the form $[x,\to)$ or $(\leftarrow,x]$ that is in $\mathscr{T}$ is in $\mathscr{T}_\preceq$. Now suppose that some $[x,y]\in\mathscr{T}$; certainly $(\leftarrow,x)$ and $(y,\to)$ are in $\mathscr{T}$, so $[x,\to)=[x,y]\cup(y,\to)\in\mathscr{T}$ and $(\leftarrow,y]=(\leftarrow,x)\cup[x,y]\in\mathscr{T}$. But then $[x,\to)\in\mathscr{T}_\preceq$ and $(\leftarrow,y]\in\mathscr{T}_\preceq$, so $[x,y]=[x,\to)\cap(\leftarrow,y]\in\mathscr{T}_\preceq$ as well. Intervals of the types $[x,y)$ and $(x,y]$ can be handled similarly, using the fact that $[x,y)=[x,\to)\cap(\leftarrow,y)$ and $(x,y]=(x,\to)\cap(\leftarrow,y]$.