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](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
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).

<b>Theorem.</b> 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 $\aleph_1$](http://mathoverflow.net/questions/78414/aleph-1-calibre/78451#78451). 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](http://en.wikipedia.org/wiki/Delta_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 Lindel&ouml;f by
the argument of [this MO question on Linedel&ouml;fness 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.

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.