Variant of strong approximation. Let $K$ be a global field. Let $w$ be a place of $K$. Let $\textbf{A}^w$ be the restricted direct product over all $v$ except $w$ of the $K_v$ with respect to the subgroups $\mathcal{O}_v$. How do I see that the diagonal image of $K^\times \to \textbf{A}^w$ is dense?
 A: Let $K$ be a global field with a distinguished place $w$.
Let $\mathbb{A}^w$ be the restricted direct product of all the $K_{v\ne w}$ with respect to the subgroups $\mathcal{O}_v$ whenever $v$ is nonarchimedean; note that these are open subsets of $K_v$, so everything is well-defined. Then $\mathbb{A}^w$ is the union of$$\prod_{v\in S} K_v \prod_{v'\notin S\sqcup\{w\}} \mathcal{O}_{v'}$$over all finite sets $S$ of places not containing $w$.
Fix a point $\textbf{a} = (a_v)_{v\ne w} \in \mathbb{A}^w$, and let $S_0$ be the finite set of nonarchimedean places $v\ne w$ where $|a_v|_v > 1$. To establish density of the diagonal image of $K^\times \hookrightarrow \mathbb{A}^w$, we need to show that each open set $U$ of $\mathbb{A}^w$ containing $\textbf{a}$ contains some $x\in K^\times \hookrightarrow \mathbb{A}^w$.
We do this by the strong approximation theorem (see here): let $U$ be such an open set containing $\textbf{a}$; we may, without loss of generality, suppose $U$ lies in the usual basis, i.e. for some finite set $T$ of places $v\ne w$ and open sets $U_v\subseteq K_v$ for these $v\in T$, we have$$U = \prod_{v\in T} U_v \prod_{v'\notin T\sqcup\{w\}} \mathcal{O}_{v'}.$$By default we are, without loss of generality, putting all finitely many archimedean places in $T$. Since $\textbf{a}\in U$, we must have $S_0 \subseteq T$, so that $a_v \in U_v$ for each $v\in T$ and $a_{v'}\in \mathcal{O}_{v'}$ for each $v'\notin T\sqcup\{w\}$.
Fix $\epsilon>0$. For each place $v\ne w$, use density of $K\hookrightarrow K_v$ to find $a_{v,n}\in K$ with $|a_{v,n} - a_v|_v < \epsilon$. By strong approximation, take $x \in K$ such that $|x - a_{v,n}|_v < \epsilon$ for all $v\in T$ and $|x|_v \le 1$ for all $v\notin T\sqcup\{w\}$; adjusting by a little — again using strong approximation, possibly being a little more careful at archimedean places — if necessary, we may assume $x \ne 0$, so $x \in K^\times$.
Then $|x - a_v|_v < 2\epsilon$ for all $v\in T$ and $|x|_{v'}\le 1$ for all $v'\notin T\sqcup\{w\}$, which are nonarchimedean by definition of $T$, automatically gives $x\in \mathcal{O}_{v'}$. Since the $U_v$ are open for $v\in T$, we may take $\epsilon>0$ small enough so that $|x - a_v|_v < 2\epsilon$ guarantees $x\in U_v$ for all $v\in T$.
So $x\in K^\times\hookrightarrow \mathbb{A}^w$ maps in diagonally as an element of $$\prod_{v\in T} U_v \prod_{v'\notin T\sqcup\{w\}} \mathcal{O}_{v'} = U,$$as desired.
