= Solution
The <integral closure> of $R$ in $T$ is
$$
\overline R^{\,T}=\{t\in T:t\text{ satisfies a monic polynomial over }R\}.
$$
It is a subring: finitely many integral elements generate a finite $R$-module algebra, and the <determinant trick> shows that each element of that algebra is integral. In particular, sums and products of integral elements remain integral.
A <valuation ring> is an <integral domain> $V$ such that, for every nonzero $z$ in its <fraction field>, either $z\in V$ or $z^{-1}\in V$. Equivalently, its <ideals> are totally ordered by inclusion. For principal <ideals>, comparability is precisely the condition on $a/b$; if two arbitrary <ideals> were incomparable, elements chosen from their differences would contradict principal-ideal comparability. Such a <ring> is local. Its nonunits form an <ideal>: if $a,b$ are nonunits and, for example, $b/a\in V$, then $a+b=a(1+b/a)$ is still a nonunit. The unique <maximal ideal> consists of those nonunits.
First, <valuation rings are integrally closed>. If $z\notin V$, then $y=z^{-1}\in V$ is a nonunit and lies in its <maximal ideal> $\mathfrak m$. A monic relation for $z$ over $V$, multiplied by $y^n$, would give
$$
1+a_{n-1}y+\cdots+a_0y^n=0,
$$
which is impossible modulo $\mathfrak m$. Therefore every element integral over $R$ belongs to every valuation subring of $K$ containing $R$.
For the reverse inclusion, we will construct a valuation overring that excludes any chosen nonintegral element. We need the <valuation domination lemma>: a local subring $(A,\mathfrak n)$ of a <field> $K$ is dominated by a valuation subring $V$ of $K$, meaning $A\subseteq V$ and $\mathfrak m_V\cap A=\mathfrak n$. Here is a proof, including the crucial maximality step.
Order the local subrings of $K$ dominating $A$ by domination. For a chain, take the union of the <rings> and of their maximal <ideals>. The union is a <local ring>: an element outside the union <ideal> is already a unit in a member of the chain, while an element in that <ideal> cannot become a unit in a later dominating member. The union still dominates $A$. Thus <Zorn's lemma> supplies a maximal pair $(V,\mathfrak m)$.
For any $z\in K^\times$, at least one of $\mathfrak mV[z]$ and $\mathfrak mV[z^{-1}]$ is proper. Suppose otherwise. There would be relations
$$
1=\sum_{i=0}^n a_i z^i,\qquad1=\sum_{j=0}^s b_j z^{-j},\qquad a_i,b_j\in\mathfrak m,
$$
with $n,s$ chosen minimal. Both are positive. Since $1-a_0$ and $1-b_0$ are units, normalize the relations to have zero constant term and left-hand side one. If $n\geq s$, the second relation gives
$$
z^s=\sum_{j=1}^s c_jz^{s-j},\qquad c_j\in\mathfrak m.
$$
Repeatedly substituting this monic reduction in the first relation yields a relation for $1$ of degree less than $s$, with every coefficient still in $\mathfrak m$. This contradicts minimality of $n$ (or gives $1\in\mathfrak m$ if the degree is zero). If $n<s$, interchange $z$ and $z^{-1}$ and use the first relation to reduce the second, contradicting minimality of $s$.
Choose whichever extension has a proper extended <ideal>, then a <maximal ideal> containing it. Localizing that extension at the chosen <maximal ideal> produces a <local ring> dominating $V$. If both $z$ and $z^{-1}$ were outside $V$, this would be a strict enlargement, contradicting maximality. Thus $V$ has the valuation property. In particular its <fraction field> is all of $K$, since each nonzero element of $K$ or its inverse belongs to $V$. This proves the <valuation domination lemma>.
Now let $x\in K$ be nonintegral over $R$, so $x\neq0$, and set $y=x^{-1}$, $B=R[y]$. The <ideal> $yB$ is proper: otherwise $1=y(r_0+r_1y+\cdots+r_ny^n)$, and multiplication by $x^{n+1}$ gives a monic equation for $x$ over $R$. Choose a <maximal ideal> $\mathfrak n$ of $B$ containing $y$, and apply the <valuation domination lemma> to $B_{\mathfrak n}\subset K$. Its dominating <valuation ring> $V$ contains $R$ and has $y\in\mathfrak m_V$. Hence $y$ is not invertible in $V$, so $x\notin V$.
We have excluded every nonintegral element from at least one valuation overring, while every integral element belongs to all of them. Therefore the <integral closure as an intersection of valuation rings> is
$$
\boxed{\overline R^{\,K}=\bigcap_{\substack{V\subseteq K\text{ a valuation ring}\\R\subseteq V}}V.}
$$
Neither <Noetherianity> nor a discrete valuation is required for this separation argument.
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