Are all compact subsets of $\mathbb R^{\mathbb R}$ separable? The title is the question. I should perhaps add that $\mathbb R$ has its usual topology and the product has the product topology, i.e., $\mathbb R^{\mathbb R}$ is the space of all functions from $\mathbb R$ to $\mathbb R$ with the topology of pointwise convergence. It is easy to see that this space is separable, e.g., the set of polynomials with rational coefficients is dense. However, in general separable topological spaces, subspaces need not be again separable.
I guess that the answer to this question is well-known -- however searching, e.g., Engelking's book for such a special question is rather frustrating.
 A: The answer is "no":
Every completely regular, T1 space of weigth $\le $ c can be embedded into $\mathbb [0,1]^{\mathbb R}$, hence into $\mathbb R^{\mathbb R}$ (see Engelking 2.3.23).
So, to find a counter-example, just take any compact, non-separable space of weigth $\le $ c, for instance the one-point-compactifaction of the discrete space of cardinality c, or the remainder of the Stone-Cech compactification of the integers.
A: For an explicit example of a compact subset that is not separable, consider the set $X$ of functions $f:\mathbb{R}\to\{0,1\}$ which are $0$ at all but at most one point.  This is a closed subset of $\{0,1\}^\mathbb{R}$ and is thus compact.  However, it is not separable, since it has uncountably many isolated points (for each $r\in\mathbb{R}$, the characteristic function of $\{r\}$ is isolated in $X$).
(In fact, this is just an explicit realization of one of the examples Ulli mentioned, namely the 1-point compactification of a discrete space of cardinality $\mathfrak{c}$.  The discrete space is the set of characteristic functions of singletons, and the point at infinity is the constant $0$ function.)
