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 $\bigcup\{U\in {\cal U}\mid U\cap Y\neq\emptyset\}=X$.
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 $\aleph_1$-calibre. Suppose that we have open sets $U_\alpha$ for $\alpha\lt\aleph_1$. By shrinking the sets, we may assume without loss of generality that each $U_\alpha$ is a basic open set. Consider the support $I_\alpha\subset\kappa$ of $U_\alpha$. If uncountably many $U_\alpha$ have the same support $J$, then since $\mathbb{R}^J$ is separable and hence $\aleph_1$-calibre, it follows that there is an uncountable subfamily with nonempty intersection. So we may assume that there are uncountably many different supports appearing for the supports of the various $U_\alpha$. Now, by the $\Delta$-system lemma, it follows that there is an uncountable subfamily $I\subset\aleph_1$, such that the supports of $U_\alpha$ for $\alpha\in I$ form a $\Delta$-system with finite root $J_0$, meaning that any two such supports intersect exactly to $J_0$. Since again $\mathbb{R}^{J_0}$ is separable, it follows again that $\mathbb{R}^{J_0}$ is $\aleph_1$-calibre and so there is an uncountable subfamily $I_0\subset I$ such that $\bigcap_{\alpha\in I_0}U_\alpha\upharpoonright J_0$ is not empty. Since it is a $\Delta$-system, it follows that $\bigcap_{\alpha\in I_0}U_\alpha$ is not empty in the original space, and so $\mathbb{R}^\kappa$ is $\aleph_1$-calibre, as desired.
Please comment if I have made a mistake. It may be that even $\mathbb{N}^\kappa$ for uncountable $\kappa$ is a counterexample, since a $\Delta$-system argument shows that it is $\aleph_1$-calibre, but 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.