TITLE: A compactness property for Borel sets
QUESTION [9 upvotes]: Is the following generalised compactness property of Borel sets in a Polish space consistent with ZFC?
($*$)  Let $\mathcal{B}$ be a family of $\aleph_1$-many Borel sets. If $\bigcap \mathcal{B} = \emptyset$, then, for some countable $\mathcal{C} \subseteq \mathcal{B}$, it holds that $\bigcap \mathcal{C} = \emptyset$.
In the special case that $\mathcal{B}$ is a family of open sets, it is not hard to show that property ($*$) is a consequence of $\neg\mathrm{CH} + \mathrm{MA}$ (where $\mathrm{CH}$ is the Continuum Hypothesis and $\mathrm{MA}$ is Martin's Axiom)—more specifically, a consequence of $\mathrm{MA}_{\aleph_1}$. Also, trivially, this special case of ($*$) itself implies $\neg \mathrm{CH}$.
However, I am stuck as to the consistency of ($*$) for general Borel sets. Actually, I have been unable to show consistency even when $\mathcal{B}$ is restricted to families of $F_\sigma$ sets.

REPLY [5 votes]: Here is a even simpler example.
Let $\{x_{\alpha} \}_{\alpha\in \omega_1}$ be an increasing chain in the Turing degrees. For every $\alpha$, let $B_{\alpha}=\{y\mid y\geq_T x_{\alpha}\}$. Each $B_{\alpha}$ is a boldface $\Sigma^0_3$ set.
Then for any countable ordinal $\beta$, $\bigcap_{\alpha<\beta}B_{\alpha}$ is not empty but $\bigcap_{\alpha<\omega_1}B_{\alpha}=\emptyset$.