$\newcommand{\cO}{\mathcal{O}}$
$\newcommand{\cS}{\mathcal{S}}$
I'm also not super experienced, so do let me know if something here is unclear or seems wrong!
As you said, we need to show that $\cO_{K,\cS}$ is an integrally closed Noetherian domain of Krull dimension (length of longest chain of prime ideals) one.
- Suppose $\alpha$ is integral over $\cO_{K,\cS}$ and satisfies the relation
$\alpha^d + a_{d-1}\alpha^{d-1} + \dots a_0 = 0$ with coefficients $a_i \in \cO_{K,\cS}$. Then, for any valuation $v$ corresponding to a prime $\mathfrak{p}$,
$$dv(\alpha) = v(a_{d-1}\alpha^{d-1} + \cdots a_0) \geq \min((d-1)v(\alpha)+v(a_{d-1}), \dots , v(a_0))$$
If $\mathfrak{p} \notin \cS$, then $v(a_i) \geq 0$ for all $i$, so it cannot be the case that $v(\alpha) < 0$. Thus, $\alpha \in \cO_{K,\cS}$, meaning that $\cO_{K,\cS}$ is integrally closed in its fraction field $K$.
As TokenToucan pointed out in the comments, the localization of a Noetherian ring is Noetherian, and $\cO_{K,S}$ is the localization with respect to the set of elements outside $\mathfrak{p}$ for all $\mathfrak{p} \notin \cS$ (if you don't see how to go from the valuation condition to ratios of elements check out this answer). Alternately, an overkill way is to observe that the Krull-Akizuki Theorem tells us that subrings of the fraction field of a one-dimensional reduced (no nonzero nilpotents) Noetherian ring are also Noetherian. In our case, $\cO_K$ is one-dimensional and an integral domain and $\cO_{K,\cS}$ contains $\cO_K$ and is contained within $K$, the fraction field of $\cO_K$.
This follows from the Krull-Akizuki theorem, as it actually guarantees that (proper) subrings of $K$ containing $\cO_K$ are Noetherian and one-dimensional.
Putting this all together, unless I messed up, we have the desired result.
The Kummer-Dedekind theorem does not require $\mathcal O_K = \mathbf Z[\alpha]$. If $K = \mathbf Q(\alpha)$ where $\alpha$ is an algebraic integer with a minimal polynomial $f(x) \in \mathbf Z[x]$, then for every prime $p$ not dividing the index $[\mathcal O_K:\mathbf Z[\alpha]]$, the factorization of $p\mathcal O_K$ resembles that of $f(x) \bmod p$. Even if you're not sure about the value of the actual index $[\mathcal O_K:\mathbf Z[\alpha]]$, that number is a factor of the very computable disciminant ${\rm disc}(f) = {\rm disc}(\mathbf Z[\alpha])$, so you can apply the theorem to all primes not dividing ${\rm disc}(f)$ while remaining unsure if it would be valid to apply it to some of the prime factors of ${\rm disc}(f)$ (settling that would require further study of the particular situation).
The reason the theorem is called "Kummer-Dedekind" is because of the role each one played, Kummer in the 1830s and Dedekind many decades later:
Kummer only treated $\mathcal O_K = \mathbf Z[\alpha]$, and that itself is being generous since he did not have a general concept of algebraic integer. He only worked with rings of integers in cyclotomic fields, where the full ring of integers very fortunately has the form $\mathbf Z[\alpha]$ and thus Kummer could succeed without realizing the true difficulty of the general case (non-monogenic rings of integers).
Dedekind came up with the general notion of algebraic integers and he proved that Kummer's theorem still works at all primes $p$ such that $p \nmid [\mathcal O_K:\mathbf Z[\alpha]]$. At first he hoped that even if $p \mid [\mathcal O_K:\mathbf Z[\alpha]]$ he could find a $\beta$ (depending on $p$) such that $p \nmid [\mathcal O_K:\mathbf Z[\beta]]$, thus allowing him to factor each $p$ by turning that into a problem of factoring a polynomial mod $p$. That hope was dashed when he discovered the first example of a non-monogenic ring of integers ($K = \mathbf Q(\alpha)$ where $\alpha^3 - \alpha^2 - 2\alpha - 8 = 0$, for which $[\mathcal O_K:\mathbf Z[\alpha]] = 2$ and in fact $2 \mid [\mathcal O_K:\mathbf Z[\beta]]$ for all $\beta \in \mathcal O_K - \mathbf Z$).
Example. $K = \mathbf Q(\sqrt[4]{24})$. The discriminant of $x^4 - 24$ is $-2^{17}3^3$, so for each prime $p$ other than $2$ and $3$, the way $(p)$ factors is like the way $x^4 - 24 \bmod p$ factors over $\mathbf F_p$. What about $p = 2$ and $p = 3$? The polynomial $x^4 - 24$ is Eisenstein at $3$, so $3 \nmid [\mathcal O_K:\mathbf Z[\sqrt[4]{24}]]$ and therefore the Kummer-Dedekind theorem says the factorization of $(3)$ resembles the factorization of $x^4 - 24 \equiv x^4 \bmod 3$, so $3$ is totally ramified in $\mathcal O_K$. To treat $p = 2$, we turn instead to $x^4 - 54$, which also has a root $\sqrt[4]{54}$ that is in $\mathcal O_K$ and generates $K$. Since $x^4 - 54$ is Eisenstein at $2$, we have $2 \nmid [\mathcal O_K:\mathbf Z[\sqrt[4]{54}]]$ and thus the Kummer-Dedekind theorem tells us that $(2)$ factors in the same way as $x^4 - 54 \equiv x^4 \bmod 54$. So $2$ is totally ramified in $\mathcal O_K$.
I am not sure why you are saying "$5$ is inert" in $\mathbf Z[\sqrt[4]{24}]$ or $\mathbf Z[\sqrt[4]{54}]$. You must have made a typographical error when coding in Sage or you misinterpreted the Sage output. Modulo $5$,
$$
x^4 - 24 = x^4 - 4 = (x^2 + 2)(x^2-2)
$$
and
$$
x^4 - 54 = x^4 - 4 = (x^2+2)(x^2-2).
$$
These are consistent with each other (both having two irreducible factors of degree $2$), as they must be since $5$ does not divide either index $[\mathcal O_K:\mathbf Z[\sqrt[4]{24}]]$ or $[\mathcal O_K:\mathbf Z[\sqrt[4]{54}]]$. Here we have an "accident" that they factor mod $5$ literally in the same way because $x^4 - 24 \equiv x^4 - 54 \bmod 5$.
In any case, the Kummer-Dedekind theorem can be used with either $x^4 - 24$ or $x^4 - 54$ to factor $(5)$ and they both tell us that $(5) = \mathfrak p\mathfrak q$ where $\mathfrak p$ and $\mathfrak q$ are prime ideals with norm $25$:
$$
\mathfrak p = (5,\sqrt[4]{24}+2) = (5,\sqrt[4]{54}-2), \ \ \ \mathfrak q = (5,\sqrt[4]{24}-2) = (5,\sqrt[4]{54}+2),
$$
where I am using positive fourth roots of $24$ and $54$ with the standard real embedding so that $\sqrt[4]{24}\sqrt[4]{54}$ is $6$ on the nose (as opposed to being $-6$).
When we declare $\mathfrak p$ to be the prime ideal $(5,\sqrt[4]{24}+2)$, modulo $\mathfrak p$ we have $\sqrt[4]{24}\sqrt[4]{54} = 6 \equiv 1 \bmod \mathfrak p$, so from $\sqrt[4]{24} \equiv -2 \equiv 3 \bmod \mathfrak p$ we must have $\sqrt[4]{54} \equiv 2 \bmod \mathfrak p$ as opposed to $\sqrt[4]{54} \equiv 3 \bmod \mathfrak p$.
I don't really need real numbers or positivity, but you have to be careful with $\sqrt[4]{24}$ and $\sqrt[4]{54}$ in
an abstract field $\mathbf Q(\gamma)$ where $\gamma^4 = 24$ because this field has two fourth roots of $24$ and two fourth roots of $54$. If you use fourth roots of both numbers together then you need to relate them to each other in a definite way to avoid making miscalculations. Since $6/\gamma$ is a fourth root of $54$, and its product with $\gamma$ is $6 \equiv 1 \bmod \mathfrak p$, we can say
$$
\mathfrak p = (5,\gamma+2) = (5,6/\gamma-2), \ \ \ \mathfrak q = (5,\gamma-2) = (5,6/\gamma+2).
$$
It would be incorrect to say $\mathfrak p = (5,\gamma+2) = (5,6/\gamma+2)$ because then modulo $\mathfrak p$ we'd have $\gamma \equiv -2 \equiv 3 \bmod \mathfrak p$ as well as $6/\gamma \equiv -2 \equiv 3 \bmod \mathfrak p$, so $\gamma \equiv 2 \bmod \mathfrak p$, and such congruences for $\gamma \bmod \mathfrak p$ (congruent to both $3$ and $2$) would be inconsistent.
Best Answer
Here is a proof that $\mathcal{O}_K/\mathcal{P}$ is finite for any nonzero prime ideal $\mathcal{P}$. Let $n$ be the degree of $K$.
First of all, for every nonzero $m \in \mathbb{Z}$, there are isomorphisms of groups $$\mathcal{O}_K/m\mathcal{O}_K \cong \mathbb{Z}^n/m\mathbb{Z}^n \cong (\mathbb{Z}/m \mathbb{Z})^n.$$ The second one is easy to see. For the first one, choose an integral basis $\beta_1,...,\beta_n$ of $K$ and write each element in $\mathcal{O}_K$ as $k_1\beta_1+\dots+k_n\beta_n$ with $k_i \in \mathbb{Z}$. Then, the association $$\mathcal{O}_K \to \mathbb{Z}^n/m\mathbb{Z}^n,\phantom{aa}k_1\beta_1+\dots+k_n\beta_n \mapsto (k_1,...,k_n) + m\mathbb{Z}^n$$ gives an onto homomorphism. It is not hard to see that $m\mathcal{O}_K$ is its kernel.
The above shows that $\mathcal{O}_K/m\mathcal{O}_K$ is finite for any nonzero $m \in \mathbb{Z}$.
Now, let's consider $\mathcal{O}_K/\mathcal{P}$. It is not hard to see that $\mathcal{P}$ contains a nonzero $m \in \mathbb{Z}$. (Take for example any nonzero $\alpha \in \mathcal{P}$ and consider its norm $N_K(\alpha)$.) Hence, $m\mathcal{O}_K \subseteq \mathcal{P}$. It follows that $$\mathcal{O}_K/m\mathcal{O}_K \to \mathcal{O}_K/\mathcal{P},\phantom{aa}\alpha+m\mathcal{O}_K \mapsto \alpha +\mathcal{P}$$ is a well defined map. It is obviously onto.
Note that exactly the same argument shows that $\mathcal{O}_K/\mathcal{I}$ is finite for any nonzero ideal $\mathcal{I}$.
For a book which contains a version of the above proof and which is also a great introduction to algebraic number theory, I recommend Number Fields by Daniel Marcus.