I read that, with the Scott topology, suprema of sequences are topological limits (See page 1 of this article).
Let $(X, \le)$ be a DCPO, and $D$ be a directed subset of $X$. I can easily see that the identity net on $D$ converges to $\bigvee D$.
How about the other direction? If the identity net on $D$ converges to $x$, does it imply that $x = \bigvee D$?
$\newcommand\LL{\mathcal L}$Brief answer: The Scott topology is not Hausdorff, and therefore we have to deal here with the set of limits, rather than with the (unique) limit. Here, for the set $\LL_D$ of limits of the identity net on $D$ we have \begin{equation*} \LL_D=L_s:=\{y\in X\colon x\le s\}, \end{equation*} where \begin{equation*} s:=s_D:=\sup D=\bigvee D. \end{equation*}
One can similarly see that the statement "Suprema of sequences are also topological limits, i.e., $\sup x_n=x \iff \lim x_n=x$" in the paper linked in your post is incorrect in general (and almost always). E.g., if $X=(-\infty,0]$ with the usual order and $x_n=0$ for all $n$, then $\sup x_n=0$, whereas the set of limits of $(x_n)$ in the Scott topology is the entire set $X=(-\infty,0]$.
Details:
First, a review of definitions.
Let $X$ be a set with a partial order $\le$.
A point $b\in X$ is an upper bound of a subset $A$ of $X$ if $x\le b$ for all $x\in A$.
A nonempty subset $D$ of $X$ is directed if any finite subset of $D$ has an upper bound in $D$.
A point $s\in X$ is the supremum of a subset $A$ of $X$ and denoted by $\sup A$ or $\bigvee A$ if $s$ is the least upper bound of $A$, that is, if $s$ is an upper bound of $A$ and for any upper bound $b$ of $A$ we have $s\le b$.
A subset $A$ of $X$ is an upper set if for any $x\in A$ we have $U_x\subseteq A$, where $U_x:=\{y\in X\colon x\le y\}$.
A subset $A$ of $X$ is a lower set if for any $x\in A$ we have $L_x\subseteq A$, where $L_x:=\{y\in X\colon y\le x\}$.
The pair $(X,\le)$ of a set $X$ with a partial order $\le$ is a directed-complete partial order (dcpo) if each directed subset $D$ of $X$ has a supremum. In what follows, $(X,\le)$ will be a dcpo.
A subset $G$ of $X$ is open in the Scott topology, or Scott-open, if $G$ is an upper set and for any directed subset $D$ of $X$ such that $\sup D\in G$ we have $D\cap G\ne\emptyset$. Equivalently, a subset $F$ of $X$ is closed in the Scott topology, or Scott-closed, if $F$ is a lower set and for any directed subset $D$ of $F$ we have $\sup D\in F$.
The identity net $I_D$ on a directed subset $D$ of $X$ is the map $D\ni x\mapsto I_D(x):=x\in X$.
The set $\LL_D$ of limits of the identity net $I_D$ is defined by the following condition on $x\in X$:
\begin{equation*} \begin{aligned} x\in\LL_D\iff &\text{for any Scott-open $G$ such that $x\in G$ } \\ &\text{there is some $z_G\in D$ such that $U_{z_G}\subseteq G$.} \end{aligned} \end{equation*}
Let now $D$ be any directed subset of $X$.
Suppose that $x\in\LL_D$. We want to show that then $x\in L_s$. Suppose the contrary. Then $x\in V:=X\setminus L_s$, and the set $V$ is Scott-open (since the set $L_x$ is Scott-closed for any $x\in X$). On the other hand, $z_V\in D$ and $s=\sup D$, so that $z_V\le s$, that is, $s\in U_{z_V}$. Also, by the definition of $\LL_D$, we have $U_{z_V}\subseteq V$. So, $s\in U_{z_V} \subseteq V=X\setminus L_s$, which contradicts the obvious fact that $s\in L_s$.
We see that $x\in L_s$ for any $x\in\LL_D$, that is, \begin{equation*} \LL_D\subseteq L_s. \tag{1}\label{1} \end{equation*}
Let us now show that $s\in\LL_D$. Take any any Scott-open $G$ such that $s\in G$. Since $G$ is Scott-open and $s=\sup D\in G$, we have $D\cap G\ne\emptyset$. Take now any $z_G\in D\cap G$. Since $z_G\in G$ and the Scott-open set $G$ is an upper set, we get $U_{z_G}\subseteq G$. Since $z_G\in D$, we see that indeed $s\in\LL_D$.
Note also that for any $x\in\LL_D$ we have $L_x\subseteq\LL_D$. Therefore and because $s\in\LL_D$, we see that $L_s\subseteq\LL_D$.
Now, in view of \eqref{1}, we conclude that \begin{equation*} \LL_D=L_s=L_{\sup D}, \end{equation*} as claimed.
