Proof.
Assumption (a) implies there is a unit $u$ of $A$ whose image in $K$ lies in the maximal ideal of $B$. Then $u$ is a nonzerodivisor of $R$ and for every $a \in A$ there exists an $n$ such that $u^ n a \in R$. It follows that $A = R_ u$.
Let $\mathfrak m_ A$ be the Jacobson radical of $A$. Let $x \in \mathfrak m_ A$ be a nonzero element. Since $\dim (A) = 1$ we see that $K = A_ x$. After replacing $x$ by $x^ n u^ m$ for some $n \geq 1$ and $m \in \mathbf{Z}$ we may assume $x$ maps to a unit of $B$. We see that for every $b \in B$ we have that $x^ nb$ in the image of $R$ for some $n$. Thus $B = R_ x$.
Let $z \in R$. If $z \not\in \mathfrak m_ A$ and $z$ does not map to an element of $\mathfrak m_ B$, then $z$ is invertible. Thus $x + u$ is invertible in $R$. Hence $\mathop{\mathrm{Spec}}(R) = D(x) \cup D(u)$. We have seen above that $D(u) = \mathop{\mathrm{Spec}}(A)$ and $D(x) = \mathop{\mathrm{Spec}}(B)$.
Case (b). If $x \in \mathfrak m_ A$, then $1 + x$ is a unit and hence $1 + x \in R$, i.e, $x \in R$. Thus we see that $\mathfrak m_ A \subset R \subset A$. In fact, in this case $A$ is integral over $R$. Namely, write $A/\mathfrak m_ A = \kappa _1 \times \ldots \times \kappa _ n$ as a product of fields. Say $x = (c_1, \ldots , c_ r, 0, \ldots , 0)$ is an element with $c_ i \not= 0$. Then
\[ x^2 - x(c_1, \ldots , c_ r, 1, \ldots , 1) = 0 \]
Since $R$ contains all units we see that $A/\mathfrak m_ A$ is integral over the image of $R$ in it, and hence $A$ is integral over $R$. It follows that $R \subset A \subset B$ as $B$ is integrally closed. Moreover, if $x \in \mathfrak m_ A$ is nonzero, then $K = A_ x = \bigcup x^{-n}A = \bigcup x^{-n}R$. Hence $x^{-1} \not\in B$, i.e., $x \in \mathfrak m_ B$. We conclude $\mathfrak m_ A \subset \mathfrak m_ B$. Thus $A \cap \mathfrak m_ B$ is a maximal ideal of $A$ thereby finishing the proof.
$\square$
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