[proofplan]
We prove that every ideal of $S^{-1}R$ is finitely generated, and then use the [ideal characterisation of Noetherian rings](/theorems/9999). If $0 \in S$, the localisation is the zero ring and the conclusion is immediate. Otherwise, given an ideal $J \trianglelefteq S^{-1}R$, we contract it to an ideal $I \trianglelefteq R$ along the localisation map, prove directly that $J$ is generated by the image of $I$, and then use finite generation of $I$ in $R$.
[/proofplan]
[step:Separate off the zero localisation case]
If $0 \in S$, then $0/1 = 1/1$ in $S^{-1}R$, so $S^{-1}R$ is the zero ring. Its only ideal is the zero ideal, which is generated by $0$, hence every ideal is finitely generated. By [citetheorem:9999], $S^{-1}R$ is Noetherian.
For the rest of the proof, assume $0 \notin S$.
[/step]
[step:Contract an arbitrary ideal of the localisation to an ideal of $R$]
Let $A := S^{-1}R$, and let
\begin{align*}
\lambda: R \to A
\end{align*}
denote the localisation homomorphism, so $\lambda(r) = r/1$ for each $r \in R$. Let $J \trianglelefteq A$ be an arbitrary ideal. Define its contraction
\begin{align*}
I := \lambda^{-1}(J) = \{r \in R : r/1 \in J\}.
\end{align*}
Because $\lambda$ is a ring homomorphism and $J$ is an ideal of $A$, the set $I$ is an ideal of $R$.
[/step]
[step:Show that the ideal in the localisation is generated by the contracted ideal]
We claim that
\begin{align*}
J = A \cdot \lambda(I),
\end{align*}
where $A \cdot \lambda(I)$ denotes the ideal of $A$ generated by the subset $\lambda(I) = \{a/1 : a \in I\}$.
First, since $I = \lambda^{-1}(J)$, every element $a \in I$ satisfies $a/1 \in J$. Hence $\lambda(I) \subset J$, and because $J$ is an ideal of $A$, we have $A \cdot \lambda(I) \subset J$.
Conversely, let $x \in J$. Since $A = S^{-1}R$, there exist $r \in R$ and $s \in S$ such that $x = r/s$. The element $s/1 \in A$ satisfies
\begin{align*}
r/1 = (s/1)(r/s).
\end{align*}
Since $r/s \in J$ and $J$ is an ideal of $A$, it follows that $r/1 \in J$. Thus $r \in I$. Therefore $r/1 \in \lambda(I)$, and
\begin{align*}
r/s = (1/s)(r/1) \in A \cdot \lambda(I).
\end{align*}
So $J \subset A \cdot \lambda(I)$, proving the equality.
[guided]
The point of passing from $J$ back to $R$ is that the Noetherian hypothesis is a statement about ideals of $R$, not directly about ideals of $S^{-1}R$. We therefore use the localisation map
\begin{align*}
\lambda: R \to S^{-1}R
\end{align*}
given by $\lambda(r) = r/1$, and define
\begin{align*}
I := \lambda^{-1}(J) = \{r \in R : r/1 \in J\}.
\end{align*}
This is an ideal of $R$ because it is the preimage of an ideal under a ring homomorphism.
We now prove, without using a full ideal-correspondence theorem, that $J$ is exactly the ideal generated by the image of $I$. Let $A := S^{-1}R$, and let $A \cdot \lambda(I)$ denote the ideal of $A$ generated by $\lambda(I) = \{a/1 : a \in I\}$.
First suppose $a \in I$. By definition of $I$, we have $a/1 \in J$. Hence every generator of $A \cdot \lambda(I)$ lies in $J$. Since $J$ is an ideal of $A$, it contains all $A$-linear combinations of these generators, so
\begin{align*}
A \cdot \lambda(I) \subset J.
\end{align*}
For the reverse inclusion, take an arbitrary element $x \in J$. Since $x \in S^{-1}R$, there are elements $r \in R$ and $s \in S$ such that $x = r/s$. The useful observation is that multiplying by $s/1$ clears the denominator:
\begin{align*}
r/1 = (s/1)(r/s).
\end{align*}
Because $r/s \in J$ and $J$ is an ideal of $A$, multiplication by the ring element $s/1 \in A$ keeps us inside $J$. Thus $r/1 \in J$, so $r \in I$ by definition of the contraction. Now $r/1$ is one of the elements used to generate $A \cdot \lambda(I)$, and multiplying it by $1/s \in A$ gives back the original element:
\begin{align*}
r/s = (1/s)(r/1) \in A \cdot \lambda(I).
\end{align*}
Since every $x \in J$ lies in $A \cdot \lambda(I)$, we have $J \subset A \cdot \lambda(I)$. Combining both inclusions gives
\begin{align*}
J = A \cdot \lambda(I).
\end{align*}
[/guided]
[/step]
[step:Use finite generation in $R$ to obtain finite generation in $S^{-1}R$]
Since $R$ is Noetherian, [citetheorem:9999] implies that every ideal of $R$ is finitely generated. Therefore the ideal $I \trianglelefteq R$ is finitely generated. Choose a finite subset $F \subset I$ such that
\begin{align*}
I = (F)_R,
\end{align*}
where $(F)_R$ denotes the ideal of $R$ generated by $F$.
We claim that $J$ is generated as an ideal of $A = S^{-1}R$ by the finite subset
\begin{align*}
\lambda(F) := \{a/1 : a \in F\}.
\end{align*}
Indeed, from the previous step,
\begin{align*}
J = A \cdot \lambda(I).
\end{align*}
For each $b \in I$, the equality $I = (F)_R$ gives a finite expression of $b$ as an $R$-linear combination of elements of $F$. Applying $\lambda$ to that expression shows that $b/1$ lies in the ideal of $A$ generated by $\lambda(F)$. Hence $A \cdot \lambda(I) = A \cdot \lambda(F)$, so
\begin{align*}
J = A \cdot \lambda(F).
\end{align*}
Thus $J$ is finitely generated.
[/step]
[step:Conclude that the localisation is Noetherian]
The ideal $J \trianglelefteq S^{-1}R$ was arbitrary, and we have shown that it is finitely generated. Hence every ideal of $S^{-1}R$ is finitely generated. By [citetheorem:9999], the ring $S^{-1}R$ is Noetherian.
[/step]
