To prove the compactness of $\mathbb{A}_{\mathbb{Q}}/ \mathbb{Q}$ (and hence $\mathbb{A}_K/K$ for an arbitrary number field $K$), one finds a set $W \subseteq \mathbb{A}_{\mathbb{Q}}$ of the form $$ \prod\limits_v \{ x \in \mathbb{Q}_v : |x|_v \leq \delta_v\}$$ where $\delta_v > 0$ is a sequence for which $\delta_v = 1$ for almost all $v$, such that $\mathbb{A}_{\mathbb{Q}} = W + \mathbb{Q}$.
To prove the strong approximation theorem, Cassels claims (Corollary on pg. 65) that the same statement holds for arbitrary $K$, i.e. there exists a similarly defined $W \subseteq \mathbb{A}_K$ such that $\mathbb{A}_K = W + K$. Why is this?
I know some authors like Serge Lang skip the reduction to $\mathbb{Q}$ and show that $\mathbb{A}_K = W + K$ holds directly, but I was hoping to see why this fact is a consequence of the case $K = \mathbb{Q}$.
I think we can argue as follows: assume we have already deduced that $\mathbb{A}_K/K$ is compact. Suppose by way of contradiction that $W + K \subsetneq \mathbb{A}_K$ for every set $W$ of the form $$\prod\limits_v \{x \in K_v : ||x_v|| \leq \delta_v\}$$ where $\delta_v$ is a sequence of positive numbers for which $\delta_v = 1$ for almost all $v$. Then $\pi(W) = \pi(W+K)$ is properly contained in $\mathbb{A}_K/K$, where $\pi$ is the quotient map $\mathbb{A}_K \rightarrow \mathbb{A}_K/K$. We can modify $W$ to make it an open set; just replace $\leq$ by $<$ when $v$ is an infinite place. Furthermore, we can find a sequence of these sets $W$, say $W_1 \subsetneq W_2 \subsetneq \cdots$ for which $$\mathbb{A}_K = \bigcup\limits_n W_n$$ by increasing the $\delta_v$, finitely many at a time. And quotient maps of topological groups are open maps (I think). We have $\mathbb{A}_K/K = \bigcup\limits_n \pi(W_n)$, so we have produced an open cover with no finite subcover.