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 < = $2^{\aleph_0}$?
The answer is negative.
According to corollary 1.34 of the paper On the extent of starcountable 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 wellknown (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 starcountable? 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. 

