Here's a cw answer so as not to leave this unanswered. If $p$ is prime, $D$ is a countable abstract set, then

(1) the product $C^D$ has exactly $\aleph_0$ closed subgroup of index $p$ (recall that for a finite index subgroup, closed=open), but

(2) it has $2^{2^{\aleph_0}}$ abstract subgroups of index $p$.

The argument for (1) is topological: this is because the Cantor set has only countably many clopen subsets.

The argument for (2) is algebraic (use that any vector space over $\mathbf{Z}/p\mathbf{Z}$ has a basis): if $R$ is an abstract set of cardinal $2^{\aleph_0}$ then $C^D$ is abstractly isomorphic to $C^{(R)}$. The set of subgroups of index $p$ is the projective space of $\text{Hom}(C^{(R)},C)=C^R$, and thus has $2^{|R|}$=$2^{2^{\aleph_0}}$ elements.

Some of these non-closed subgroups can, in a sense, be made explicit by considering an ultrafilter and consider the kernel of the natural map from the product to the ultraproduct. This is of course not really explicit since the ultrafilter itself cannot be made explicit. On the other hand, this latter argument works for an arbitrary nontrivial finite group $C$.

noin general. E.g. consider an infinite product of non-trivial copies of $\mathbb F_p$ (the finite field with $p$ elements). This will be compact, and the closed subgroups of index $p$ will be precisely the kernels of the continuous projections onto $\mathbb F_p$. But there are many linear functionals on this space that are not continuous, and there kernels will be index $p$ subgroups that are not closed. Regards, $\endgroup$ – Emerton Jun 5 '13 at 22:33