If a topological space X has $\aleph_1$-calibre, then it must be star countable? If a topological space X has $\aleph_1$-calibre[definition], then it must be star countable?
What if the cardinality of the topological space X is additionally < = $2^{\aleph_0}$?
 A: The answer is negative.


*

*A space $X$ is star countable if for every open cover
$\cal U$, there is a countable subset $Y\subset X$ such
that $\bigcup\{U\in {\cal U}\mid U\cap
Y\neq\emptyset\}=X$. 

*The space $X$ has calibre $\aleph_1$ if for every uncountable list of nonempty open sets $U_\alpha$ for $\alpha\lt\omega_1$, there is an uncountable sublist $I_0\subset \omega_1$, such that $\bigcap_{\alpha\in I_0}U_\alpha\neq\emptyset$. 
According to corollary 1.34 of the paper On the extent of
star-countable
spaces,
it is claimed that $\mathbb{R}^\kappa$ is not star
countable for sufficiently large cardinals $\kappa$.
Meanwhile, let's argue that $\mathbb{R}^\kappa$ has
calibre $\aleph_1$. In fact, it suffices to argue that the class of spaces with calibre $\aleph_1$ is closed under arbitrary products. And I have now revised my answer to give the proof of this more general fact, which Henno says in the comments is well-known (and he evidently runs in quality circles).
Theorem. The product $\Pi_{i\in I}X_i$ of any family of spaces $X_i$ with calibre $\aleph_1$ has calibre $\aleph_1$. 
Proof. First, note that finite products of calibre $\aleph_1$ spaces has calibre $\aleph_1$ by the argument given in this MO question on finite products of calibre $\aleph_1$. Suppose that $U_\alpha$ for $\alpha\lt\omega_1$ is an uncountable family of open sets in the product. By shrinking the sets,
we may assume without loss of generality that each
$U_\alpha$ is a basic open set, having some finite support $I_\alpha\subset I$. If uncountably many
$U_\alpha$ have the same finite support $J$, then by since $\prod_{i\in J}X_i$ is a finite product and hence has calibre $\aleph_1$, 
it follows that there is an uncountable subfamily of these $U_\alpha$ with
nonempty intersection, witnessing this instance of calibre $\aleph_1$ for the product. So we are left with the case where there are
uncountably many different supports appearing for the
supports of the various $U_\alpha$. Thus, we have an uncountable family of finite sets $I_\alpha$. By the
$\Delta$-system lemma, there is an
uncountable subfamily $I_0\subset \omega_1$, such that the
supports of $U_\alpha$ for $\alpha\in I_0$ form a
$\Delta$-system with finite root $J_0$, meaning that any
two such supports intersect exactly to $J_0$. Since again
$\prod_{i\in J_0}X_i$ has calibre $\aleph_1$, it follows that there is an
uncountable subfamily $I_1\subset I_0$ such that
$\bigcap_{\alpha\in I_1}U_\alpha\upharpoonright J_0$ is not
empty. Since it is a $\Delta$-system, it also follows that
$\bigcap_{\alpha\in I_1}U_\alpha$ is not empty in the
original space, since the remaining parts of the supports do not conflict with each other, and so $\prod_{i\in I}X_i$ has calibre $\aleph_1$, as desired. QED
In particular, $\mathbb{R}^\kappa$ has calibre $\aleph_1$, but is not star countable, thereby answering the first (original) question. 
Perhaps $\mathbb{N}^\kappa$ for uncountable $\kappa$ may be a simpler 
counterexample, since the theorem also shows it to have calibre $\aleph_1$, and we know at least that it is not Lindelöf by
the argument of this MO question on Linedelöfness and
compactness.
Is it star-countable? I'm not currently sure, but if not,
then it may be a simpler counterexample.
But perhaps one needs $\kappa$ to be very large to make this conclusion, in which case even $\mathbb{N}^\kappa$ may not answer the second question. 
