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I'll prove the theorem claimed by Kunen, assuming basic knowledge about elementary sub-models.

If $(Y, \tau)$ is a compact Hausdorff space and $\mathcal{F}$ is a cover of $Y$ by closed $G_\delta$ sets and $\mathcal{F}$ satisfies: for every $H \in \mathcal{F}$ the set $\{K \in \mathcal{F}: K \cap H \neq \emptyset \}$ has cardinality at most continuum then $|\mathcal{F}| \leq 2^\omega$.

Let $\theta$ be a large enough regular cardinal and $M \prec H(\theta)$ be a countably closed elementary submodel of $H(\theta)$ of cardinality $\mathfrak{c}$ such that $\mathfrak{c} \subset M$ and $\{Y, \tau, \mathcal{F}, \mathfrak{c}\} \subset M$.

We claim that $\mathcal{F} \subset M$. Suppose this is not the case and let $K \in \mathcal{F} \setminus M$.

Using compactness and regularity of $Y$, we can write every closed $G_\delta$ set $F \subset Y$ in the form $F=\bigcap \{\overline{V_n(F)}: n < \omega \}$, for a suitable decreasing family of open neighbourhoods $\{V_n(F): n < \omega \}$ of $F$.

If $F \in \mathcal{F} \cap M$, then the family $\{V_n(F): n < \omega \}$ may be fixed in $M$, but then we actually have $\{V_n(F): n < \omega \} \subset M$.

Note now that for every $x \in \overline{Y \cap M}$, we can find $F \in \mathcal{F} \cap M$ such that $x \in F$. Indeed, let $x \in \overline{Y \cap M} \setminus Y \cap M$. Then there is $F \in \mathcal{F}$ such that $x \in F$. Let $x_n$ be any point of $U_n(F) \cap Y \cap M$. Using compactness of $F$ and the fact that $F=\bigcap \{\overline{U_n(F)}: n <\omega \}$ we can easily see that the sequence $\{x_n: n < \omega \}$ converges to $F$, in the sense that, every open neighbourhood of $F$ contains a final segment of $\{x_n: n < \omega \}$. Now by $\omega$-closedness of $M$, we have that $\{x_i: i \geq n \} \in M$, for every $n<\omega$ and hence the set:

$$\mathcal{S}=\{K \in \mathcal{F}: \{x_n: n <\omega \} \rightarrow K \}$$

is an element of $M$ (the formula defining that set has all free variables in $M$).

Note that since every compact Hausdorff space is normal, $F$ can be separated from a closed set disjoint from it by a pair of disjoint open sets. Therefore, we have that if $G \in \mathcal{S}$ then $G \cap F \neq \emptyset$. Hence $|\mathcal{S}| \leq 2^{\aleph_0}$ by assumption. But then we actually have that $\mathcal{S} \subset M$ and hence $F \in M$, as we wanted.

For every $F \in \mathcal{F} \cap M$ we have $\{H \in \mathcal{F} : F \cap H \neq \emptyset \} \in M$, and, since $|\{H \in \mathcal{F} : F \cap H \neq \emptyset \}| \leq \mathfrak{c}$, we also have $\{H \in \mathcal{F} : F \cap H \neq \emptyset \} \subset M$. So, for every $F \in \mathcal{F} \cap M$, we have $F \cap K=\emptyset$.

Now, by compactness of $K$, we can find for every $F \in \mathcal{F} \cap M$, an integer $n <\omega$ such that $\overline{U_n(F)} \cap K=\emptyset$.

Thus for every $x \in \overline{Y \cap M}$, we can find an open neighbourhood $U_x \in M$ of $x$ such that $U_x \cap K=\emptyset$.

Now $\mathcal{U}=\{U_x: x \in \overline{Y \cap M} \}$ is an open cover of the compact space $\overline{Y \cap M}$ such that $\mathcal{U} \subset M$. So there is a finite set $\mathcal{F} \subset \mathcal{U}$ such that $\overline{Y \cap M} \subset \bigcup \mathcal{F}$. This implies that $M \models Y \subset \bigcup \mathcal{F}$ and hence $H(\theta) \models Y \subset \bigcup \mathcal{F}$. But this contradicts $K \cap \bigcup \mathcal{F}=\emptyset$.

So $\mathcal{F} \subset M$ and hence $|\mathcal{F}| \leq |M| \leq \mathfrak{c}$.

I'll prove the theorem claimed by Kunen, assuming basic knowledge about elementary sub-models.

If $(Y, \tau)$ is a compact Hausdorff space and $\mathcal{F}$ is a cover of $Y$ by closed $G_\delta$ sets and $\mathcal{F}$ satisfies: for every $H \in \mathcal{F}$ the set $\{K \in \mathcal{F}: K \cap H \neq \emptyset \}$ has cardinality at most continuum then $|\mathcal{F}| \leq 2^\omega$.

Let $\theta$ be a large enough regular cardinal and $M \prec H(\theta)$ be a countably closed elementary submodel of $H(\theta)$ of cardinality $\mathfrak{c}$ such that $\mathfrak{c} \subset M$ and $\{Y, \tau, \mathcal{F}, \mathfrak{c}\} \subset M$.

We claim that $\mathcal{F} \subset M$. Suppose this is not the case and let $K \in \mathcal{F} \setminus M$.

Using compactness and regularity of $Y$, we can write every closed $G_\delta$ set $F \subset Y$ in the form $F=\bigcap \{\overline{V_n(F)}: n < \omega \}$, for a suitable decreasing family of open neighbourhoods $\{V_n(F): n < \omega \}$ of $F$.

If $F \in \mathcal{F} \cap M$, then the family $\{V_n(F): n < \omega \}$ may be fixed in $M$, but then we actually have $\{V_n(F): n < \omega \} \subset M$.

Note now that for every $x \in \overline{Y \cap M}$, we can find $F \in \mathcal{F} \cap M$ such that $x \in F$. Indeed, let $x \in \overline{Y \cap M} \setminus Y \cap M$. Then there is $F \in \mathcal{F}$ such that $x \in F$. Let $x_n$ be any point of $U_n(F) \cap Y \cap M$. Using compactness of $F$ and the fact that $F=\bigcap \{\overline{U_n(F)}: n <\omega \}$ we can easily see that the sequence $\{x_n: n < \omega \}$ converges to $F$, in the sense that, every open neighbourhood of $F$ contains a final segment of $\{x_n: n < \omega \}$. Now by $\omega$-closedness of $M$, we have that $\{x_i: i \geq n \} \in M$, for every $n<\omega$ and hence the set:

$$\mathcal{S}=\{K \in \mathcal{F}: \{x_n: n <\omega \} \rightarrow K \}$$

is an element of $M$ (the formula defining that set has all free variables in $M$).

Note that since every compact Hausdorff space is normal, $F$ can be separated from a closed set disjoint from it by a pair of disjoint open sets. Therefore, we have that if $G \in \mathcal{S}$ then $G \cap F \neq \emptyset$. Hence $|\mathcal{S}| \leq 2^{\aleph_0}$ by assumption. But then we actually have that $\mathcal{S} \subset M$ and hence $F \in M$, as we wanted.

For every $F \in \mathcal{F} \cap M$ we have $\{H \in \mathcal{F} : F \cap H \neq \emptyset \} \in M$, and, since $|\{H \in \mathcal{F} : F \cap H \neq \emptyset \}| \leq \mathfrak{c}$, we also have $\{H \in \mathcal{F} : F \cap H \neq \emptyset \} \subset M$. So, for every $F \in \mathcal{F} \cap M$, we have $F \cap K=\emptyset$.

Now, by compactness of $K$, we can find for every $F \in \mathcal{F} \cap M$, an integer $n <\omega$ such that $\overline{U_n(F)} \cap K=\emptyset$.

Thus for every $x \in \overline{Y \cap M}$, we can find an open neighbourhood $U_x \in M$ of $x$ such that $U_x \cap K=\emptyset$.

Now $\mathcal{U}=\{U_x: x \in \overline{Y \cap M} \}$ is an open cover of the compact space $\overline{Y \cap M}$ such that $\mathcal{U} \subset M$. So there is a finite set $\mathcal{G} \subset \mathcal{U}$ such that $\overline{Y \cap M} \subset \bigcup \mathcal{G}$. This implies that $M \models Y \subset \bigcup \mathcal{G}$ and hence $H(\theta) \models Y \subset \bigcup \mathcal{G}$. But this contradicts $K \cap \bigcup \mathcal{G}=\emptyset$.

So $\mathcal{F} \subset M$ and hence $|\mathcal{F}| \leq |M| \leq \mathfrak{c}$.

I'll prove the theorem claimed by Kunen, assuming basic knowledge about elementary sub-models.

If $Y$ is a compact Hausdorff space and $\mathcal{F}$ is a cover of $Y$ by closed $G_\delta$ sets and $\mathcal{F}$ satisfies: for every $H \in \mathcal{F}$ the set $\{K \in \mathcal{F}: K \cap H \neq \emptyset \}$ has cardinality at most continuum then $|\mathcal{F}| \leq 2^\omega$.

Let $\theta$ be a large enough regular cardinal and $M \prec H(\theta)$ be a countably closed elementary submodel of $H(\theta)$ of cardinality $\mathfrak{c}$ that contains everything we need ($Y$, $\mathcal{F}$, etc...).

We claim that $\mathcal{F} \subset M$. Suppose this is not the case and let $K \in \mathcal{F} \setminus M$.

Using compactness and regularity of $Y$, we can write every closed $G_\delta$ set $F \subset Y$ in the form $F=\bigcap \{\overline{V_n(F)}: n < \omega \}$, for a suitable decreasing family of open neighbourhoods $\{V_n(F): n < \omega \}$ of $F$.

If $F \in \mathcal{F} \cap M$, then the family $\{V_n(F): n < \omega \}$ may be fixed in $M$, but then we actually have $\{V_n(F): n < \omega \} \subset M$.

Note now that for every $x \in \overline{Y \cap M}$, we can find $F \in \mathcal{F} \cap M$ such that $x \in F$. Indeed, let $x \in \overline{Y \cap M} \setminus Y \cap M$. Then there is $F \in \mathcal{F}$ such that $x \in F$. Let $x_n$ be any point of $U_n(F) \cap Y \cap M$. Using compactness of $F$ and the fact that $F=\bigcap \{\overline{U_n(F)}: n <\omega \}$ we can easily see that the sequence $\{x_n: n < \omega \}$ converges to $F$, in the sense that, every open neighbourhood of $F$ contains a final segment of $\{x_n: n < \omega \}$. Now by $\omega$-closedness of $M$, we have that $\{x_i: i \geq n \} \in M$, for every $n<\omega$ and hence the set:

$$\mathcal{S}=\{K \in \mathcal{F}: \{x_n: n <\omega \} \rightarrow K \}$$

is an element of $M$ (the formula defining that set has all free variables in $M$).

Note that since every compact Hausdorff space is normal, $F$ can be separated from a closed set disjoint from it by a pair of disjoint open sets. Therefore, we have that if $G \in \mathcal{S}$ then $G \cap F \neq \emptyset$. Hence $|\mathcal{S}| \leq 2^{\aleph_0}$ by assumption. But then we actually have that $\mathcal{S} \subset M$ and hence $F \in M$, as we wanted.

For every $F \in \mathcal{F} \cap M$ we have $\{H \in \mathcal{F} : F \cap H \neq \emptyset \} \in M$, and, since $|\{H \in \mathcal{F} : F \cap H \neq \emptyset \}| \leq \mathfrak{c}$, $M \prec H(\theta)$ and $|M| = \mathfrak{c}$, we also have $\{H \in \mathcal{F} : F \cap H \neq \emptyset \} \subset M$. So, for every $F \in \mathcal{F} \cap M$, we have $F \cap K=\emptyset$.

Now, by compactness of $K$, we can find for every $F \in \mathcal{F} \cap M$, an integer $n <\omega$ such that $\overline{U_n(F)} \cap K=\emptyset$.

Thus for every $x \in \overline{Y \cap M}$, we can find an open neighbourhood $U_x \in M$ of $x$ such that $U_x \cap K=\emptyset$.

Now $\mathcal{U}=\{U_x: x \in \overline{Y \cap M} \}$ is an open cover of the compact space $\overline{Y \cap M}$ such that $\mathcal{U} \subset M$. So there is a finite set $\mathcal{F} \subset \mathcal{U}$ such that $\overline{Y \cap M} \subset \bigcup \mathcal{F}$. This implies that $M \models Y \subset \bigcup \mathcal{F}$ and hence $H(\theta) \models Y \subset \bigcup \mathcal{F}$. But this contradicts $K \cap \bigcup \mathcal{F}=\emptyset$.

So $\mathcal{F} \subset M$ and hence $|\mathcal{F}| \leq |M| \leq \mathfrak{c}$.

I'll prove the theorem claimed by Kunen, assuming basic knowledge about elementary sub-models.

If $(Y, \tau)$ is a compact Hausdorff space and $\mathcal{F}$ is a cover of $Y$ by closed $G_\delta$ sets and $\mathcal{F}$ satisfies: for every $H \in \mathcal{F}$ the set $\{K \in \mathcal{F}: K \cap H \neq \emptyset \}$ has cardinality at most continuum then $|\mathcal{F}| \leq 2^\omega$.

Let $\theta$ be a large enough regular cardinal and $M \prec H(\theta)$ be a countably closed elementary submodel of $H(\theta)$ of cardinality $\mathfrak{c}$ such that $\mathfrak{c} \subset M$ and $\{Y, \tau, \mathcal{F}, \mathfrak{c}\} \subset M$.

We claim that $\mathcal{F} \subset M$. Suppose this is not the case and let $K \in \mathcal{F} \setminus M$.

Using compactness and regularity of $Y$, we can write every closed $G_\delta$ set $F \subset Y$ in the form $F=\bigcap \{\overline{V_n(F)}: n < \omega \}$, for a suitable decreasing family of open neighbourhoods $\{V_n(F): n < \omega \}$ of $F$.

If $F \in \mathcal{F} \cap M$, then the family $\{V_n(F): n < \omega \}$ may be fixed in $M$, but then we actually have $\{V_n(F): n < \omega \} \subset M$.

Note now that for every $x \in \overline{Y \cap M}$, we can find $F \in \mathcal{F} \cap M$ such that $x \in F$. Indeed, let $x \in \overline{Y \cap M} \setminus Y \cap M$. Then there is $F \in \mathcal{F}$ such that $x \in F$. Let $x_n$ be any point of $U_n(F) \cap Y \cap M$. Using compactness of $F$ and the fact that $F=\bigcap \{\overline{U_n(F)}: n <\omega \}$ we can easily see that the sequence $\{x_n: n < \omega \}$ converges to $F$, in the sense that, every open neighbourhood of $F$ contains a final segment of $\{x_n: n < \omega \}$. Now by $\omega$-closedness of $M$, we have that $\{x_i: i \geq n \} \in M$, for every $n<\omega$ and hence the set:

$$\mathcal{S}=\{K \in \mathcal{F}: \{x_n: n <\omega \} \rightarrow K \}$$

is an element of $M$ (the formula defining that set has all free variables in $M$).

Note that since every compact Hausdorff space is normal, $F$ can be separated from a closed set disjoint from it by a pair of disjoint open sets. Therefore, we have that if $G \in \mathcal{S}$ then $G \cap F \neq \emptyset$. Hence $|\mathcal{S}| \leq 2^{\aleph_0}$ by assumption. But then we actually have that $\mathcal{S} \subset M$ and hence $F \in M$, as we wanted.

For every $F \in \mathcal{F} \cap M$ we have $\{H \in \mathcal{F} : F \cap H \neq \emptyset \} \in M$, and, since $|\{H \in \mathcal{F} : F \cap H \neq \emptyset \}| \leq \mathfrak{c}$, we also have $\{H \in \mathcal{F} : F \cap H \neq \emptyset \} \subset M$. So, for every $F \in \mathcal{F} \cap M$, we have $F \cap K=\emptyset$.

Now, by compactness of $K$, we can find for every $F \in \mathcal{F} \cap M$, an integer $n <\omega$ such that $\overline{U_n(F)} \cap K=\emptyset$.

Thus for every $x \in \overline{Y \cap M}$, we can find an open neighbourhood $U_x \in M$ of $x$ such that $U_x \cap K=\emptyset$.

Now $\mathcal{U}=\{U_x: x \in \overline{Y \cap M} \}$ is an open cover of the compact space $\overline{Y \cap M}$ such that $\mathcal{U} \subset M$. So there is a finite set $\mathcal{F} \subset \mathcal{U}$ such that $\overline{Y \cap M} \subset \bigcup \mathcal{F}$. This implies that $M \models Y \subset \bigcup \mathcal{F}$ and hence $H(\theta) \models Y \subset \bigcup \mathcal{F}$. But this contradicts $K \cap \bigcup \mathcal{F}=\emptyset$.

So $\mathcal{F} \subset M$ and hence $|\mathcal{F}| \leq |M| \leq \mathfrak{c}$.

I'll prove the theorem claimed by Kunen, assuming basic knowledge about elementary sub-models.

If $Y$ is a compact Hausdorff space and $\mathcal{F}$ is a cover of $Y$ by closed $G_\delta$ sets and $\mathcal{F}$ satisfies: for every $H \in \mathcal{F}$ the set $\{K \in \mathcal{F}: K \cap H \neq \emptyset \}$ has cardinality at most continuum then $|\mathcal{F}| \leq 2^\omega$.

Let $\theta$ be a large enough regular cardinal and $M \prec H(\theta)$ be a countably closed elementary submodel of $H(\theta)$ of cardinality $\mathfrak{c}$ that contains everything we need ($Y$, $\mathcal{F}$, etc...).

We claim that $\mathcal{F} \subset M$. Suppose this is not the case and let $K \in \mathcal{F} \setminus M$.

Using compactness and regularity of $Y$, we can write every closed $G_\delta$ set $F \subset Y$ in the form $F=\bigcap \{\overline{V_n(F)}: n < \omega \}$, for a suitable decreasing family of open neighbourhoods $\{V_n(F): n < \omega \}$ of $F$.

If $F \in \mathcal{F} \cap M$, then the family $\{V_n(F): n < \omega \}$ may be fixed in $M$, but then we actually have $\{V_n(F): n < \omega \} \subset M$.

Note now that for every $x \in \overline{Y \cap M}$, we can find $F \in \mathcal{F} \cap M$ such that $x \in F$. Indeed, let $x \in \overline{Y \cap M} \setminus Y \cap M$. Then there is $F \in \mathcal{F}$ such that $x \in F$. Let $x_n$ be any point of $U_n(F) \cap Y \cap M$. Using compactness of $F$ and the fact that $F=\bigcap \{\overline{U_n(F)}: n <\omega \}$ we can easily see that the sequence $\{x_n: n < \omega \}$ converges to $F$, in the sense that, every open neighbourhood of $F$ contains a final segment of $\{x_n: n < \omega \}$. Now by $\omega$-closedness of $M$, we have that $\{x_i: i \geq n \} \in M$, for every $n<\omega$ and hence the set:

$$\mathcal{S}=\{K \in \mathcal{F}: \{x_n: n <\omega \} \rightarrow K \}$$

is an element of $M$ (the formula defining that set has all free variables in $M$).

Note that since every compact Hausdorff space is normal, $F$ can be separated from a closed set disjoint from it by a pair of disjoint open sets. Therefore, we have that if $G \in \mathcal{S}$ then $G \cap F \neq \emptyset$. Hence $|\mathcal{S}| \leq 2^{\aleph_0}$ by assumption. But then we actually have that $\mathcal{S} \subset M$ and hence $F \in M$, as we wanted.

By assumption, for every $F \in \mathcal{F} \cap M$ we have $\{H \in \mathcal{F} : F \cap H \neq \emptyset \} \in M$, and, since $|\{H \in \mathcal{F} : F \cap H \neq \emptyset \}| \leq \mathfrak{c}$, $M \prec H(\theta)$ and $|M| = \mathfrak{c}$, we also have $\{H \in \mathcal{F} : F \cap H \neq \emptyset \} \subset M$. So, for every $F \in \mathcal{F} \cap M$, we have $F \cap K=\emptyset$.

Now, by compactness of $K$, we can find for every $F \in \mathcal{F} \cap M$, an integer $n <\omega$ such that $\overline{U_n(F)} \cap K=\emptyset$.

Thus for every $x \in \overline{Y \cap M}$, we can find an open neighbourhood $U_x \in M$ of $x$ such that $U_x \cap K=\emptyset$.

Now $\mathcal{U}=\{U_x: x \in \overline{Y \cap M} \}$ is an open cover of the compact space $\overline{Y \cap M}$ such that $\mathcal{U} \subset M$. So there is a finite set $\mathcal{F} \subset \mathcal{U}$ such that $\overline{Y \cap M} \subset \bigcup \mathcal{F}$. This implies that $M \models Y \subset \bigcup \mathcal{F}$ and hence $H(\theta) \models Y \subset \bigcup \mathcal{F}$. But this contradicts $K \cap \bigcup \mathcal{F}=\emptyset$.

So $\mathcal{F} \subset M$ and hence $|\mathcal{F}| \leq |M| \leq \mathfrak{c}$.

I'll prove the theorem claimed by Kunen, assuming basic knowledge about elementary sub-models.

If $Y$ is a compact Hausdorff space and $\mathcal{F}$ is a cover of $Y$ by closed $G_\delta$ sets and $\mathcal{F}$ satisfies: for every $H \in \mathcal{F}$ the set $\{K \in \mathcal{F}: K \cap H \neq \emptyset \}$ has cardinality at most continuum then $|\mathcal{F}| \leq 2^\omega$.

Let $\theta$ be a large enough regular cardinal and $M \prec H(\theta)$ be a countably closed elementary submodel of $H(\theta)$ of cardinality $\mathfrak{c}$ that contains everything we need ($Y$, $\mathcal{F}$, etc...).

We claim that $\mathcal{F} \subset M$. Suppose this is not the case and let $K \in \mathcal{F} \setminus M$.

Using compactness and regularity of $Y$, we can write every closed $G_\delta$ set $F \subset Y$ in the form $F=\bigcap \{\overline{V_n(F)}: n < \omega \}$, for a suitable decreasing family of open neighbourhoods $\{V_n(F): n < \omega \}$ of $F$.

If $F \in \mathcal{F} \cap M$, then the family $\{V_n(F): n < \omega \}$ may be fixed in $M$, but then we actually have $\{V_n(F): n < \omega \} \subset M$.

Note now that for every $x \in \overline{Y \cap M}$, we can find $F \in \mathcal{F} \cap M$ such that $x \in F$. Indeed, let $x \in \overline{Y \cap M} \setminus Y \cap M$. Then there is $F \in \mathcal{F}$ such that $x \in F$. Let $x_n$ be any point of $U_n(F) \cap Y \cap M$. Using compactness of $F$ and the fact that $F=\bigcap \{\overline{U_n(F)}: n <\omega \}$ we can easily see that the sequence $\{x_n: n < \omega \}$ converges to $F$, in the sense that, every open neighbourhood of $F$ contains a final segment of $\{x_n: n < \omega \}$. Now by $\omega$-closedness of $M$, we have that $\{x_i: i \geq n \} \in M$, for every $n<\omega$ and hence the set:

$$\mathcal{S}=\{K \in \mathcal{F}: \{x_n: n <\omega \} \rightarrow K \}$$

is an element of $M$ (the formula defining that set has all free variables in $M$).

Note that since every compact Hausdorff space is normal, $F$ can be separated from a closed set disjoint from it by a pair of disjoint open sets. Therefore, we have that if $G \in \mathcal{S}$ then $G \cap F \neq \emptyset$. Hence $|\mathcal{S}| \leq 2^{\aleph_0}$ by assumption. But then we actually have that $\mathcal{S} \subset M$ and hence $F \in M$, as we wanted.

For every $F \in \mathcal{F} \cap M$ we have $\{H \in \mathcal{F} : F \cap H \neq \emptyset \} \in M$, and, since $|\{H \in \mathcal{F} : F \cap H \neq \emptyset \}| \leq \mathfrak{c}$, $M \prec H(\theta)$ and $|M| = \mathfrak{c}$, we also have $\{H \in \mathcal{F} : F \cap H \neq \emptyset \} \subset M$. So, for every $F \in \mathcal{F} \cap M$, we have $F \cap K=\emptyset$.

Now, by compactness of $K$, we can find for every $F \in \mathcal{F} \cap M$, an integer $n <\omega$ such that $\overline{U_n(F)} \cap K=\emptyset$.

Thus for every $x \in \overline{Y \cap M}$, we can find an open neighbourhood $U_x \in M$ of $x$ such that $U_x \cap K=\emptyset$.

Now $\mathcal{U}=\{U_x: x \in \overline{Y \cap M} \}$ is an open cover of the compact space $\overline{Y \cap M}$ such that $\mathcal{U} \subset M$. So there is a finite set $\mathcal{F} \subset \mathcal{U}$ such that $\overline{Y \cap M} \subset \bigcup \mathcal{F}$. This implies that $M \models Y \subset \bigcup \mathcal{F}$ and hence $H(\theta) \models Y \subset \bigcup \mathcal{F}$. But this contradicts $K \cap \bigcup \mathcal{F}=\emptyset$.

So $\mathcal{F} \subset M$ and hence $|\mathcal{F}| \leq |M| \leq \mathfrak{c}$.