The main question in this post is: How to proof the Chinese remainder theorem (in elementary number theory, i.e. in $\mathbb{Z}$) using the strong approximation theorem in $\mathbb{Q}$ in valuation theory.
Any proof and references are welcomed! :)
We shall state the strong approximation theorem here. It is clearer to introduce the weak approximation theorem at first:
Weak approximation theorem: Given $n$ inequivalent nontrivial valuation $\vert \cdot \vert_i$, $i=1,\ldots,n$ of a field $k$, an arbitrary positive real number $\epsilon$, and $n$ arbitrary elements $a_i$, there exists an element $a \in k$ such that $$ \vert a - a_i \vert_i < \epsilon.$$
The strong approximation theorem of $\mathbb{Q}$ goes like this (keep using the notations above):
Strong approximation theorem: Let $k$ above be the rational number field $\mathbb{Q}$, and the valuations $\vert \cdot \vert_i$ be $p_i$-adic valuations, then not only does there exists an $a \in \mathbb{Q}$ such that $ \vert a - a_i \vert_i < \epsilon$ for $i=1,\ldots,n$, but $ \vert a\vert_p \leq 1$ for all other $p$-adic valuations as well.
The source of this question and my attempts: I have heard people said that the approximation theorem in valuation theory is somehow a generalization of Chinese remainder theorem. So I am wondering whether we can use the approximation theorem to prove the Chinese remainder theorem. By looking up many books, especially on G. Bachman’s Introduction to $p$-adic numbers and valuation theory, there is an exercise asking for proof of the Chinese remainder theorem using the strong approximation theorem.
The Chinese Remainder Theorem has various equivalent fomulations, but let's take this one:
That such an $a$, if it exists, is unique modulo $lcm(n_1, ..., n_k) =n$ is easy to show by elementary means. The main thing to prove, for which we can use the strong approximation theorem, is existence of $a$.
Now notice, following user Berci's comment, that in the special case that each $n_i$ is a prime power $p_i^{k_i}$, this is almost literally the theorem as quoted by you, applied with $\epsilon := \min_i \{p_i^{-k_i}\}$. Namely, the $a \in \mathbb Q$ which now exists due to that theorem actually is an integer $a \in \mathbb Z$ (that is what $\lvert a \rvert_p \le 1$ for all primes $p$ means), and for each $i$, $\lvert a-a_i\rvert < \epsilon \le p^{-k_i}$ literally means that $a \equiv a_i$ (mod $p^\ell$) for some $\ell > k_i$ which is even stronger than $a \equiv a_i$ (mod $p^k_i$).
In the general case, we decompose each $n_i$ into its prime (power) factors; formally, let's say we have numerated all primes $p_1, p_2, p_3, ...$, then for each $i$ let $J(i)$ be the set $\{j \in \mathbb N: p_j \mid n_i \}$ of those primes which divide $n_i$, so that $n_i = \displaystyle\prod_{j \in J(i)} p_{j}^{k_{j}}$. Note that since the $n_i$ are mutually coprime, the sets $J(i)$ are mutually disjoint; let $J := \bigcup_i J(i)$ be their union and $a_j := a_i$ for all $j \in J(i)$. Now apply the theorem to the $a_j, j \in J$ and $\epsilon:= \min_{j \in J} \{p_j^{-k_j}\}$. Check that again the $a$ whose existence comes from the approximation theorem is an integer such that for each $i$, the difference $a-a_i$ is divisible by all $p_j^{k_j}$ for $j \in J(i)$, hence by their product $n_i$, in other words $a \equiv a_i$ (mod $n_i$).