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](http://www.cs.vu.nl/~vanmill/papers/papers2011/al-ju-vm-tk-wi.pdf),
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](http://mathoverflow.net/questions/9641/how-far-is-lindelof-from-compactness/9651#9651).
Is it star-countable? I'm not currently sure, but if not,
then it may be a simpler counterexample.