When $F=\mathbb{Q}$ and we consider open sets defined using only the finite places, this says that given finitely many primes $p_1,\dots,p_n$, integers $N_1,\dots,N_n$, and elements $x_i\in\mathbb{Q}_{p_i}$, there exists $x\in\mathbb{Q}$ such that $\nu_{p_i}(x-x_i)\geq N_i$ for each $i$. Scaling everything by appropriate powers of the $p_i$ and adjusting the $N_i$ as appropriate, we may assume that the $x_i$ are all in $\mathbb{Z}_{p_i}$. We may also assume $N_i\geq 0$ for all $i$ so that this forces the denominator of $x$ to not be divisible by any $p_i$.

So, then, what do we have? We have an element $x\in\mathbb{Z}[S^{-1}]$ where $S$ is all the primes except $p_1,\dots,p_n$, such that $x-x_i$ is divisible by $p_i^{N_i}$ for each $i$. Or, reducing mod $m=p_1^{N_1}\dots p_n^{N_n}$, this says exactly that the natural map $\mathbb{Z}[S^{-1}]/(m)\to \prod_i\mathbb{Z}_{p_i}/(p_i^{N_i})$ is surjective. Since every element of $S$ is a unit mod $m$, $\mathbb{Z}[S^{-1}]/(m)$ can be identified with $\mathbb{Z}/(m)$, and $\mathbb{Z}_{p_i}/(p_i^{N_i})$ can be identified with $\mathbb{Z}/(p_i^{N_i})$, so this says that the natural map $\mathbb{Z}/(m)\to\prod_i\mathbb{Z}/(p_i^{N_i})$ is surjective. That's the Chinese remainder theorem.