Family of sets closed under arbitrary intersections and arbitrary unions of chains of its elements induces a finitary closure operator? Let $\mathcal{U} \subseteq \wp\left ( X \right )$ be a family of sets that contains both $\emptyset$ and $X$, is closed under arbitrary intersections and is closed under arbitrary unions of chains of its elements under inclusion. Formally, let $\mathcal{U}$ satisfy the following:

*

*$\emptyset, X \in \mathcal{U}$

*$ \left \{U_i  \right \}_{i \in I} \subseteq \mathcal{U} \Rightarrow \bigcap_{i \in I}^{} U_i \in \mathcal{U}$

*$\left \{ U_i \right \}_{i \in I} \subseteq \mathcal{U} : \forall \alpha, \beta \in I : U_{\alpha}\subseteq U_{\beta} \vee U_{\beta} \subseteq U_{\alpha} \Rightarrow \bigcup_{i \in I}^{} U_i \in \mathcal{U}$
We define a closure operator  for every $S \subseteq X$ as:
$\mathrm{cl}\left ( S \right )=\bigcap \left \{U : U\supseteq S \quad \mathrm{and} \quad U \in \mathcal{U} \right \}$
How can I prove that $\mathrm{cl}: \mathcal{P}(X) \rightarrow \mathcal{P}(X)$ is a finitary closure operator? That is, how can I prove that for every $S \subseteq X$ the following is satisfied?
$\mathrm{cl} \left ( S \right )=\bigcup \left \{\mathrm{cl}(U) : U\subseteq S \quad \mathrm{and} \quad U \;
 \mathrm{finite} \right \}$
This result is (without proof) stated in https://math.berkeley.edu/~gbergman/245/2.4/Ch.5.pdf as Lemma 5.3.6.
 A: Recall that a family of subsets $\mathcal{F} \subseteq \mathcal{P}(X)$ is called directed if for any $F_1, F_2 \in \mathcal{F}$ there is $F_3 \in \mathcal{F}$ that contains both $F_1$ and $F_2$.
Fact. Any $\mathcal{U} \subseteq \mathcal{P}(X)$ is closed under unions of directed families iff it is closed under unions of chains.
See here for a question about exactly this statement, with a proof in the answers. It also appears in Locally Presentable and Accessible Categories by Adámek and Rosický, Lemma 1.6.
Now we note that $\mathcal{F} = \{ \operatorname{cl}(U) : U \subseteq S \text{ and } U \text{ is finite}\}$ is a directed family. Indeed, for $\operatorname{cl}(U_1), \operatorname{cl}(U_2) \in \mathcal{F}$ we have that $\operatorname{cl}(U_1 \cup U_2) \in \mathcal{F}$ contains both of them. By assumption $\mathcal{U}$ is closed under unions of chains, so by the fact it is closed under unions of directed families. So $\bigcup \mathcal{F} \in \mathcal{U}$ and as clearly $S \subseteq \bigcup \mathcal{F}$ we then have $\operatorname{cl}(S) \subseteq \bigcup \mathcal{F}$. The other inclusion, $\bigcup \mathcal{F} \subseteq \operatorname{cl}(S)$ is easy and already holds without the assumption about unions of chains.
