I'm thinking about results of the form: Under assumption $A$, if $X$ is a compact Hausdorff space and $C$ is a cover of $X$ by closed $G_\delta$ sets, then there is a subcover of cardinality $\leq\kappa$.

Obviously in general we can take $\kappa\leq|X|$ and in some cases this is the best we can do such as in Cantor space in which every point is a closed $G_\delta$ set.

The strongest result I've found so far is this: If $X$ is a scattered compact Hausdorff space and $C$ is a cover of $X$ by closed $G_\delta$ sets, then there is a countable subcover.

Proof: Since $X$ is scattered it is totally disconnected, and we can represent each $F\in C$ as $F=\bigcap_{n<\omega}Q_{F,n}$, where $Q_{F,n}$ is a nested sequence of clopen sets. Partition $X$ into a finite collection $X_0, \dots ,X_{n-1}$ of clopen sets such that each $X_i$ has a unique point $x_i$ such that $CB(x_i) = CB(X)$ (where $CB$ is the Cantor-Bendixson rank). In each $X_i$ choose some $F_i\in C$ such that $x_i \in F$ and then consider $X_i \setminus F_i$, which can be written as a countable union of clopen sets. Iterate this procedure in each such clopen set. This forms a well-founded forest which is countably branching and thus countable. The subcover is the $F_i$'s chosen at each node of the forest.

This is certainly optimal for the general scattered case, as you can construct a cover that saturates the bound in any infinite scattered compact Hausdorff space.

A weak generalization is this: If $X$ is a compact Hausdorff space with perfect core $P$, and $C$ is a cover of $X$ by closed $G_\delta$ sets, then for any $C_0 \subseteq C$ covering $P$, $C$ has a subcover of cardinality $\leq \omega^{|C_0| + 1}$.

Proof: If $P$ is empty we can use the previous result, so assume that $P$ is non-empty. Let $C_0$ be a cover of $P$, so in particular $P \subseteq \bigcup_{F\in C_0}\bigcap_{n<\omega}U_{F,n} = \bigcap_{g:C_0\rightarrow\omega}\bigcup_{F\in C_0}U_{F,g(F)}$. By compactness, for each $g:C_0 \rightarrow \omega$, there is a finite set $C_{0,g}\subseteq C_0$ such that $P\subseteq \bigcup_{F\in C_{0,g}}U_{F,g(F)}$. So let $R_g=X\setminus \bigcup_{F\in C_{0,g}}U_{F,g(F)}$. Each $R_g$ is closed and scattered as a subspace of $X$, so we can run the previous argument to get a countable $C_g \subseteq C$ that covers $R_g$. This means that in order to cover the rest of $X$ we need at most $\omega^{|C_0|}\times \omega = \omega^{|C_0|+1}$ more sets, so since $\omega^{|C_0|+1} \geq |C_0|$ we need at most $\omega^{|C_0|+1}$ all together.

This is only clearly non-trivial in cases where $P$ is much smaller than $X$. The question is how much better can you do? In particular is there a universal bound for all compact Hausdorff spaces? Can you get a better bound in terms of $|C_0|$? Can you get bounds in terms of other cardinal invariants of $P$ or $X$, such as the weight or density character? Can you say anything non-trivial about perfect spaces? What is $\kappa$ for $\beta \omega \setminus \omega$ or $2^\lambda$ for various cardinals $\lambda$?

EDIT: In the negative direction I suspect you can show that if $X$ is a compact Hausdorff space with non-empty perfect core, then $X$ has a cover by closed $G_\delta$ sets such that no subcover has cardinality $< 2^\omega$.

EDIT2: Suppose that $X$ is a compact Hausdorff space with non-empty perfect core $P$. Construct a tree of closed sets $\{F_{\sigma}\}_{\sigma \in 2^{<\omega}}$ with the following properties: Every $F^\circ_{\sigma}$ has non-empty intersection with $P$. For every $\sigma \in 2^{<\omega}$, $F_{\sigma\frown 0},F_{\sigma\frown 1} \subset F_\sigma^\circ$ and $F_{\sigma\frown 0}\cap F_{\sigma \frown 1} = \varnothing$. (We can do this since $P$ is perfect.) By these conditions, for any $\alpha \in 2^\omega$, the set $G_\alpha = \bigcap_{n<\omega} F_{\alpha \upharpoonright n}$ is a closed $G_\delta$ set. If $\alpha,\beta \in 2^\omega$ with $\alpha \neq \beta$, then $G_\alpha \cap G_\beta = \varnothing$, and $H = \bigcup_{\alpha \in 2^\omega} G_\alpha$ is a closed $G_\delta$ set. Since $H$ is a closed $G_\delta$ set, we can find a sequence of closed $G_\delta$ sets $\{D_n\}_{n<\omega}$ such that $\bigcup_{n<\omega} D_n = X \setminus H$. So then the cover $\{D_n\}_{n<\omega} \cup \{ G_\alpha \}_{\alpha \in 2^\omega}$ is a cover by closed $G_\delta$ sets with no subcover of cardinality $< 2^\omega$.

EDIT3: There's a much better bound in the second case that I missed: If $X$ is a compact Hausdorff space with perfect core $P$, and $C$ is a cover of $X$ by closed $G_\delta$ sets, then for any $C_0 \subseteq C$ covering $P$, $C$ has a subcover of cardinality $\leq |C_0|+\omega$.

Proof: Same proof as before, just notice that while there are $\omega^{|C_0|}$ many functions $g:C_0 \rightarrow \omega$, there are only $(|C_0|\times \omega)^{<\omega} = |C_0|+\omega$ many finite subsets of $\{U_{F,n}\}_{F\in C,n<\omega}$, so there are only at most $|C_0|+\omega$ many $R_g$'s to consider, and we get that $C$ has a subcover of size at most $|C_0|+(|C_0|+\omega)\times\omega = |C_0|+\omega$.

By the same argument as in the scattered case, this is basically optimal relative to the size of $|C_0|$, so now the questions really is about perfect spaces.

EDIT4: Combining the content of Taras and Anonymous's comments, you get a more refined bound in terms of generalized Cantor-Bendixson ranks (these probably have a name somewhere), specifically for any cardinal $\kappa$ we can can define

$$X^\prime_{\kappa}=X \setminus \bigcup \{U\subseteq X : U\text{ open, has density character }\leq\kappa\}$$

$$X^{(\alpha)}_{\kappa} = \bigcap_{\beta < \alpha} (X^{(\beta)}_{\kappa})^\prime _{\kappa}$$

where $\bigcap\varnothing=X$, for any $\alpha \in \text{Ord}\cup\{\infty\}$. And $CB_{\kappa}(X)$ is the smallest ordinal $\alpha$ for which $X^{(\alpha + 1)}_{\kappa}=\varnothing$, if it exists, and $\infty$ otherwise. Then if we let $S(X)$ be the smallest $\kappa$ such that $CB_{\kappa}(X)<\infty$, we should get a better bound in terms of $S(X)$.

If the density character of $X$ is $\kappa$, then every closed $G_\delta$ cover has a subcover of cardinality $\leq 2^\kappa$, since there are only at most that many closed $G_\delta$ sets in $X$.

Assume that for some cardinal $\kappa$ and some ordinal $\alpha$, we've shown that if $CB_{\kappa}(X)<\alpha$, then every closed $G_\delta$ cover has a subcover of cardinality $\leq 2^\lambda$ for some $\lambda < \kappa$. Let $X$ be a compact Hausdorff space such that $CB_{\kappa}(X) = \alpha$ and let $C$ be a cover of $X$ by closed $G_\delta$ sets. This implies that the density character of $P = X^{(CB_{\kappa}(X))}_{\kappa}$ is $\leq\kappa$, so there is some set $C_0 \subseteq C$ of cardinality $\leq 2^\kappa$ such that $C_0$ covers $P$. By the same argument as before, $X\setminus \bigcup C_0$ can be written as the union of at most $2^\kappa$ many closed subspaces, each of which has strictly smaller $CB_{\kappa}$. So by the induction hypothesis each of these has a subset of $C$ of cardinality $\leq 2^\kappa$ that covers it, so overall $X$ is covered by a subset of $C$ of cardinality $\leq 2^\kappa$.

So by induction, if $CB_{\kappa}(X)<\infty$, then every closed $G_\delta$ cover of $X$ has a subcover of cardinality $\leq 2^\kappa$. So in general any closed $G_\delta$ cover of a compact Hausdorff space has a subcover of cardinality $\leq 2^{S(X)}$.

You can probably get slightly tighter bounds at singular cardinals using a CB derivative for density characters $<\kappa$ rather than just $\leq \kappa$.