 Solution

Underthe preimage of a basic product neighbourhood $U\times V$ isAt almost all finite places, $U_v=V_v={\mathcal{O}}_v$, and the condition $x_v,x_v^{-1}\in {\mathcal{O}}_v$ is exactly $x_v\in {\mathcal{O}}_v^{\times }$. Thus these preimages form the usual restricted-product basis for ${\mathbb{I}}_K$. Conversely, every basic idele neighbourhood is obtained by choosing suitable $U_v$ and $V_v$ locally. Hence the restricted product topology on the idele group equals the subspace topology induced by $\varphi$.