Show that the definition of Scott topology is a topology

Question

To verify the definition of a Scott topology is a topology, I still need to show that it's closed under intersection. Can someone help?

Definition 1 (Scott topology). Let $(D,\leq)$ be a complete partial order. The Scott topology on $D$ is defined as follows, $$O\subseteq D$$ is open if

1. $x\in O \wedge x\leq y \implies y\in O$
2. $\sup X\in O, X\subseteq D$ directed $\implies X\cap O \neq \emptyset$

Definition 2 (Directed). A subset $X\subseteq D$ is directed if $X\neq \emptyset$ and $$\forall x,y\in X \ \exists z\in X \, (x\leq z \land y\leq z)$$

Answer 1

Proving that Scott-open sets are closed under finite intersection amounts to prove that, given some Scott-open sets $O_1, \dots, O_n \subseteq D$ with $n \geq 0$, their intersection $O = O_1 \cap \dots \cap O_n$ is still Scott-open. Let us prove it, according to the definition of Scott-open.

0. Clearly, $O \subseteq D$ (note that if $n = 0$ then $O = D$, see here).

1. Let $x \in O$ and $x \leq y \in D$. For all $1 \leq i \leq n$, from $x \in O$ it follows that $x \in O_i$, and so (since $O_i$ is Scott-open) $y \in O_i$. Hence, $y \in O$.

2. Let $X \subseteq D$ be directed with $\sup X \in O$. For all $1 \leq i \leq n$, from $\sup X \in O$ it follows that $\sup X \in O_i$, and so (since $O_i$ is Scott-open) $X \cap O_i \neq \emptyset$; let $x_i \in X \cap O_i$. I claim that $ X \cap O \neq \emptyset$. For simplicity's sake, consider the case $n = 2$, and $x_1 \in X \cap O_1$ and $x_2 \in X \cap O_2$. Since $X$ is directed, there is a $x \in X$ such that $x_1 \leq x$ and $x_2 \leq x$. Since $O_1$ and $O_2$ are Scott-open, $x \in O_1$ and $x \in O_2$. Thus, $x \in X \cap O_1 \cap O_2$ and so $X \cap O = X \cap O_1 \cap O_2 \neq \emptyset$. The argument is similar for any $n \geq 0$, formally it is a proof by induction on $n$.