Let $V_i\subset K$ be a family of valuation rings (not necessarily discrete) associated to the field $K$. Set $D=\bigcap_iV_i$, where index $i$ may run through an infinite set. Is $\mathrm{Frac}(D)=K$?
For finite intersection case, the book has shown that $\mathrm{Frac}(D)=K$ as each $V_i$ is a localization of $D$ at a maximal ideal. However, consider $Z=\cap_{p}Z_p$ where index $p$ runs through all primes of $Z$. Then it seems plausible that $Frac(D)=K$ still. I have to note that $Z,Z_p$ are noetherian but there are non-noetherian valuation rings.
Ref. Nagata, Local Rings Chpt 1, Sec 11.
Take the example $K=\mathbb{C}(X)$, and consider the discrete $\mathbb{C}$-valuations, which are given by the $(X-a)$-adic valuation $v_a$ for each $a\in \mathbb{C}$, and the valuation at infinity $v_\infty$ which is just minus the degree.
Then the intersection of the valuation rings associated to all $v_a$ consists of the rational fractions having no pole, which means it is $\mathbb{C}[X]$. But then the intersection with the valuation associated to $v_\infty$ is just $\mathbb{C}$ since polynomials have nonnegative degree.
So the total intersection is just $\mathbb{C}$, and obviously does not have $\mathbb{C}(X)$ as its fraction field.