For the first claim: The subset of the finite ideles given by $\prod_v \mathcal{O}_{K_v}^\times$ is open by the definition of the topology on the restricted direct product, and an element in $K^\times$ is in this subset iff it is a unit in each completion, which is true iff it's in $\mathcal{O}^\times$.
The second claim is just a generalization of this first argument. A basis of open neighborhoods of 1 in the finite ideles is given by picking any open subgroup you choose of $K_v^\times$ in finitely many of the $v$-spots and picking $\mathcal{O}_v^\times$ in all the rest of them. If we just want a basis of open subsets, we may as well only look at really small ones, so we can assume that in the finitely many $v$-spots where we didn't pick $\mathcal{O}_v^\times$, we picked some small open subset of $\mathcal{O}_v^\times$. Note that subgroups of $\mathcal{O}_v^\times$ are all finite index and generated by a power of the uniformizer. Note also that all these open subsets of the ideles are subsets of the one we discussed in checking the first claim.
What happens when we intersect such a neighborhood with the diagonal image of $K^\times$? We get the subgroup of elements in $\mathcal{O}_K^\times$ which are in all the open subgroups of $\mathcal{O}_v^\times$ that we picked at the special places $v$. The index is just the product of the indices of those subgroups, so that's a finite index subgroup. Conversely, any finite index subgroup of $\mathcal{O}_K^\times$ is determined by the index of its image in each completion.