If $A$ is a Krull ring of $K$, then $A \cap K'$ is also a Krull ring for $K' \subset K$. 
If $A$ is a Krull ring of $K$, then $A \cap K'$ is also a Krull ring for $K' \subset K$ (where $K'$ is a subfield of $K$).

What's confusing me is that $K'$ may not contain the uniformising element $t$ of $R_{\lambda}$ where $R_{\lambda}$ is in the defining family of $A$. How do we know that the valuation is still discrete?

Consider the restriction to $K'$ of all discrete valuations defining the Krull domain $A$. These are discrete or improper. Now get rid the improper ones and you are done. 
