$\newcommand{\U}{\mathcal{U}}$
$\newcommand{\F}{\mathcal{F}}$
$\newcommand{\D}{\mathcal{D}}$
$\newcommand{\C}{\mathcal{C}}$
For any infinite $X \subseteq \omega$, we define:
$$
\D_X := \{Y \in [\omega]^\omega : Y \subseteq X \vee Y \cap X = \emptyset\}
$$
It's easy to see that $\D_X$ is dense open for all infinite $X$. Now let $\cal{C} \subseteq [\omega]^\omega$ be a collection of infinite sets. Consider the following statement:
> The statement $\mathsf{U}(\C)$ asserts that: If $\F$ is a filter on $\omega$ and $\F \cap \D_X \neq \emptyset$ for all $X \in \cal{C}$, then $\F$ is an ultrafilter.

It's easy to verify that that $\mathsf{U}([\omega]^\omega)$ is true. Let $\mathfrak{u}'$ be the least cardinal such that there exists some $\cal{C} \subseteq [\omega]^\omega$ which $|\C| = \mathfrak{u}'$ and $\mathsf{U}(\C)$ holds. My questions are:

1. Does $\mathsf{ZFC}$ prove that $\mathfrak{u}' = \mathfrak{u}$, where $\mathfrak{u}$ is the [ultrafilter number](https://en.wikipedia.org/wiki/Cardinal_characteristic_of_the_continuum#Ultrafilter_number_%7F'%22%60UNIQ--postMath-00000045-QINU%60%22'%7F)?

2. If the answer to (1) is no, then is it consistent that $\mathfrak{u}' < \mathfrak{c}$?