Is every semi-stratifiable space $\omega$-monolithic? Is every semi-stratifiable space $\omega$-monolithic?

Definitions
A topological space $(X,\tau)$ is called semi-stratifiable if there exists a function $g:\omega\times X\to\tau$ such that:


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*for any point $x$ of $X$ holds $\{x\}=\bigcap_{n\in\omega} g(n,x)$;

*for any point $x$ of $X$ and a sequence $\{x_n\}$ of $X$, if $x \in g(n,x_n)$ for each $n$, then $x_n \to x$.
A topological space $X$ is said to be $\omega$-monolithic if $nw(\overline{A}) \le \omega$ for any subset $A \subset X$ with $|A| \le \omega$.
$nw(X)$ denotes the cardinal function
 called network weight, which is minimal cardinality of a network
$$nw(X)=\min\{|\mathcal N|: \mathcal N \text{ is a net for } X\}+\omega.$$
If not. What if $X$ is semi-metric space?
Note that $X$ is semi-metrisable iff $X$ is first countable and semi-stratifiable;
 A: As a counterexample to this question we can consider the Katetov extension $\kappa\omega$ of the  discrete space of all finite ordinals $\omega$. 
By definition, $\kappa\omega$ is the space of all ultrafilters on $\omega$ with the topology in which a neighborhood base of an ultrafilter $\mathcal U$ consists of the sets $\{\mathcal U\}\cup U$ where $U\in\mathcal U$. Here we identify $\omega$ with the set of principal ultrafilters on $\omega$. So, $\kappa\omega=\omega\cup\omega^*$ where $\omega^*$ is the set of free ultrafilters on $\omega$. The space $\kappa\omega$ is separable but has cardinality $2^{\mathfrak c}>\mathfrak c$. Since the subspace $\omega^*$ of free ultrafilters is discrete and uncountable, the separable space $\kappa\omega$ has uncountable network weight, so is not $\omega$-monolithic.
On the other hand, the space $\kappa\omega$ is semi-stratifiable. This is witnessed by the function $g$ defined by $g(n,\mathcal U)=\{\mathcal U\}$ if the ultrafilter $\mathcal U$ is principal and $g(n,\mathcal U)=\{\mathcal U\}\cup(\omega\setminus n)$ if $\mathcal U$ is free. 
This example also answers A question on semi-stratifiable space.
