Compact Hausdorff Spaces with pre-caliber $\aleph_1$ has caliber $\aleph_1$ Let us recall that a topological space has $\aleph_1$ pre-caliber (resp. caliber) if given any family of open sets $\{U_\alpha\}_{\alpha<\omega_1}$ there exists an uncontable set $B\subset\omega_1$ such that the subfamily $\{U_\alpha\}_{\alpha\in B}$ has the finite intersection property (resp. $\bigcap_{\alpha\in B} U_\alpha\neq\emptyset$).
I was trying to prove that a compact Hausdorff space has $\aleph_1$ pre-caliber if and only if it has $\aleph_1$ caliber. In this regard, I would construct a subfamily of closed sets with the FIP inside the family $\{U_\alpha\}_{\alpha\in B}$ and then using the compactness ensure that $\bigcap_{\alpha\in B} U_\alpha\neq\emptyset$. Probably using the normality of the space $X$ we can construct that subfamily but I'm not sure.
Can anybody help me? Thanks in advance.
 A: Let $F=\{U_\alpha :\alpha\in \omega_1\}$ be an open family with $\varnothing\not \in F$. For $\alpha\in \omega_1,$ let $V_\alpha$ be open with $\varnothing\ne V_\alpha\subset \overline {V_\alpha}\subset U_\alpha$. Let $B$ be an uncountable subset of $\omega_1$ such that $\{V_\alpha :\alpha\in B\}$ has the F.I.P. Then $\{\overline V_\alpha :\alpha\in B\}$ has the F.I.P., so by compactness we have $\varnothing\ne \bigcap_{\alpha\in B}\overline {V_\alpha}\subset \bigcap_{\alpha\in B}U_\alpha$.
A: I think I have a possible answer. Let us consider $\{U_\alpha\}_{\alpha<\omega_1}$ an uncountable family of open sets. From this family we can consider an uncountable subfamily $\{U_\alpha\}_{\alpha\in B}$ with the FIP due to $X$ has $\aleph_1$ as a pre-caliber. 
Now, given any finite set $F\in [B]^{<\omega}$, let us consider an open set $V_F$ such that $\overline{V_F}\subset \bigcap_{\alpha\in F} U_\alpha$ (this is possible by regularity and by the FIP). Now, since $X$ has pre-caliber $\aleph_1$ we can pick from $\{V_F\}_{F\in [B]^{<\omega}}$ another time an uncontable family $\{V_G\}_{G\in\mathcal{G}}$ of open sets with the FIP such that $\overline{V_G}\subset \bigcap_{\alpha\in G} U_\alpha$ for every $G\in \mathcal{G}$. So now it's enough to consider as a subfamily of our initial family $\{U_\alpha\}_{\alpha<\omega_1}$ the collection $\{U_\alpha\}_{\alpha\in \mathcal{A}}$ where $\mathcal{A}$ is $\bigcup \mathcal{G}$ because 
$$\emptyset \neq \bigcap_{G\in\mathcal{G}} \overline{V_G}\subset \bigcap_{\alpha\in\mathcal{A}} U_\alpha$$
by compactness.
