Let $K$ be a number field and $C_K$ be its idele class group. Exercise 5.11 in Milne's notes asks me to show that there is a finite-index subgroup $H$ of $C_K$ which is not open. I haven't found an example, and am having trouble believing it for the following reason:
If $K=\mathbb{Q}$, there is an isomorphism of topological groups
$C_\mathbb{Q} \cong \mathbb{R}^\times_{>0} \times \hat{\mathbb{Z}}^\times$.
Since $\mathbb{R}^\times_{>0}$ has no nontrivial finite-index subgroups, $H$ must contain it, so $H \mathbin/ \mathbb{R}^\times_{>0}$ is a finite-index subgroup of $\hat{\mathbb{Z}}^\times$. However, finite-index implies open for the latter.
What am I missing?
Contrary to the statement in the OP, the profinite group $\widehat{\mathbb Z}^{\times}$ admits non-open finite index subgroups.
For example, it admits a surjection onto $\prod_{n = 1}^{\infty} \mathbb Z/2\mathbb Z$, and this product has non-open finite index subgroups. (They correspond to non-principal ultrafilters on $\mathbb N$.)