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David E Speyer
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All the previous answers send us to complicated examples since these are $1$-dimensional domains. But the OP has looked for

A domain $D$ all of whose localizations $D_P$ for $P∈\mathrm{Spec}(D)$ are noetherian, and $D$ is not noetherian.

A simple example is the following: $D=\mathbb Z[\frac Xp:p\text{ prime},p\ge 2].$
This is an integral domain which is not noetherian.
Now let $P$ be a prime ideal of $D$. There are two cases:
$\bullet$ $P\cap\mathbb Z=(0)$. Set $S=D\setminus\{0\}$$S=\mathbb{Z} \setminus\{0\}$. Then $D_P\simeq(S^{-1}D)_{S^{-1}P}$, that is, $D_P$ is a localization of $S^{-1}D=\mathbb Q[X]$ which is a noetherian ring.
$\bullet$ $P\cap\mathbb Z=q\mathbb Z$ with $q\ge2$ a prime number. Set $S=\mathbb Z\setminus q\mathbb Z$. Analogously $D_P$ is a localization of $S^{-1}D=\mathbb Z_{(q)}[\frac Xq]$ (here $\mathbb Z_{(q)}$ stands for the localization of $\mathbb Z$ at the prime ideal $q\mathbb Z$) which is also a noetherian ring.

Remark. The foregoing shows that $\dim D=2$.

All the previous answers send us to complicated examples since these are $1$-dimensional domains. But the OP has looked for

A domain $D$ all of whose localizations $D_P$ for $P∈\mathrm{Spec}(D)$ are noetherian, and $D$ is not noetherian.

A simple example is the following: $D=\mathbb Z[\frac Xp:p\text{ prime},p\ge 2].$
This is an integral domain which is not noetherian.
Now let $P$ be a prime ideal of $D$. There are two cases:
$\bullet$ $P\cap\mathbb Z=(0)$. Set $S=D\setminus\{0\}$. Then $D_P\simeq(S^{-1}D)_{S^{-1}P}$, that is, $D_P$ is a localization of $S^{-1}D=\mathbb Q[X]$ which is a noetherian ring.
$\bullet$ $P\cap\mathbb Z=q\mathbb Z$ with $q\ge2$ a prime number. Set $S=\mathbb Z\setminus q\mathbb Z$. Analogously $D_P$ is a localization of $S^{-1}D=\mathbb Z_{(q)}[\frac Xq]$ (here $\mathbb Z_{(q)}$ stands for the localization of $\mathbb Z$ at the prime ideal $q\mathbb Z$) which is also a noetherian ring.

Remark. The foregoing shows that $\dim D=2$.

All the previous answers send us to complicated examples since these are $1$-dimensional domains. But the OP has looked for

A domain $D$ all of whose localizations $D_P$ for $P∈\mathrm{Spec}(D)$ are noetherian, and $D$ is not noetherian.

A simple example is the following: $D=\mathbb Z[\frac Xp:p\text{ prime},p\ge 2].$
This is an integral domain which is not noetherian.
Now let $P$ be a prime ideal of $D$. There are two cases:
$\bullet$ $P\cap\mathbb Z=(0)$. Set $S=\mathbb{Z} \setminus\{0\}$. Then $D_P\simeq(S^{-1}D)_{S^{-1}P}$, that is, $D_P$ is a localization of $S^{-1}D=\mathbb Q[X]$ which is a noetherian ring.
$\bullet$ $P\cap\mathbb Z=q\mathbb Z$ with $q\ge2$ a prime number. Set $S=\mathbb Z\setminus q\mathbb Z$. Analogously $D_P$ is a localization of $S^{-1}D=\mathbb Z_{(q)}[\frac Xq]$ (here $\mathbb Z_{(q)}$ stands for the localization of $\mathbb Z$ at the prime ideal $q\mathbb Z$) which is also a noetherian ring.

Remark. The foregoing shows that $\dim D=2$.

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user26857
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All the previous answers send us to complicated examples since these are $1$-dimensional domains. But the OP has looked for

A domain $D$ all of whose localizations $D_P$ for $P∈\mathrm{Spec}(D)$ are noetherian, and $D$ is not noetherian.

A simple example is the following: $D=\mathbb Z[\frac Xp:p\text{ prime},p\ge 2].$
This is an integral domain which is not noetherian: $(\frac{X}{p_1})\subsetneq(\frac{X}{p_1p_2})\subsetneq\cdots$.
Now let $P$ be a prime ideal of $D$. There are two cases:
$\bullet$ $P\cap\mathbb Z=(0)$. Set $S=D\setminus\{0\}$. Then $D_P\simeq(S^{-1}D)_{S^{-1}P}$, that is, $D_P$ is a localization of $S^{-1}D=\mathbb Q[X]$ which is a noetherian ring.
$\bullet$ $P\cap\mathbb Z=q\mathbb Z$ with $q\ge2$ a prime number. Set $S=\mathbb Z\setminus q\mathbb Z$. Analogously $D_P$ is a localization of $S^{-1}D=\mathbb Z_{(q)}[\frac Xq]$ (here $\mathbb Z_{(q)}$ stands for the localization of $\mathbb Z$ at the prime ideal $q\mathbb Z$) which is also a noetherian ring.

Remark. The foregoing shows that $\dim D=2$.

All the previous answers send us to complicated examples since these are $1$-dimensional domains. But the OP has looked for

A domain $D$ all of whose localizations $D_P$ for $P∈\mathrm{Spec}(D)$ are noetherian, and $D$ is not noetherian.

A simple example is the following: $D=\mathbb Z[\frac Xp:p\text{ prime},p\ge 2].$
This is an integral domain which is not noetherian: $(\frac{X}{p_1})\subsetneq(\frac{X}{p_1p_2})\subsetneq\cdots$.
Now let $P$ be a prime ideal of $D$. There are two cases:
$\bullet$ $P\cap\mathbb Z=(0)$. Set $S=D\setminus\{0\}$. Then $D_P\simeq(S^{-1}D)_{S^{-1}P}$, that is, $D_P$ is a localization of $S^{-1}D=\mathbb Q[X]$ which is a noetherian ring.
$\bullet$ $P\cap\mathbb Z=q\mathbb Z$ with $q\ge2$ a prime number. Set $S=\mathbb Z\setminus q\mathbb Z$. Analogously $D_P$ is a localization of $S^{-1}D=\mathbb Z_{(q)}[\frac Xq]$ (here $\mathbb Z_{(q)}$ stands for the localization of $\mathbb Z$ at the prime ideal $q\mathbb Z$) which is also a noetherian ring.

Remark. The foregoing shows that $\dim D=2$.

All the previous answers send us to complicated examples since these are $1$-dimensional domains. But the OP has looked for

A domain $D$ all of whose localizations $D_P$ for $P∈\mathrm{Spec}(D)$ are noetherian, and $D$ is not noetherian.

A simple example is the following: $D=\mathbb Z[\frac Xp:p\text{ prime},p\ge 2].$
This is an integral domain which is not noetherian.
Now let $P$ be a prime ideal of $D$. There are two cases:
$\bullet$ $P\cap\mathbb Z=(0)$. Set $S=D\setminus\{0\}$. Then $D_P\simeq(S^{-1}D)_{S^{-1}P}$, that is, $D_P$ is a localization of $S^{-1}D=\mathbb Q[X]$ which is a noetherian ring.
$\bullet$ $P\cap\mathbb Z=q\mathbb Z$ with $q\ge2$ a prime number. Set $S=\mathbb Z\setminus q\mathbb Z$. Analogously $D_P$ is a localization of $S^{-1}D=\mathbb Z_{(q)}[\frac Xq]$ (here $\mathbb Z_{(q)}$ stands for the localization of $\mathbb Z$ at the prime ideal $q\mathbb Z$) which is also a noetherian ring.

Remark. The foregoing shows that $\dim D=2$.

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user26857
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All the previous answers send us to complicated examples since these wereare $1$-dimensional domains. But the OP has looked for

A domain $D$ all of whose localizations $D_P$ for $P∈\mathrm{Spec}(D)$ are noetherian, and $D$ is not noetherian.

A simple example is the following: $D=\mathbb Z[\frac Xp:p\text{ prime},p\ge 2].$
This is an integral domain which is not noetherian: $(\frac{X}{p_1})\subsetneq(\frac{X}{p_1p_2})\subsetneq\cdots$.
Now let $P$ be a prime ideal of $D$. There are two cases:
$\bullet$ $P\cap\mathbb Z=(0)$. Set $S=D\setminus\{0\}$. Then $D_P\simeq(S^{-1}D)_{S^{-1}P}$, that is, $D_P$ is a localization of $S^{-1}D=\mathbb Q[X]$ which is a noetherian ring.
$\bullet$ $P\cap\mathbb Z=q\mathbb Z$ with $q\ge2$ a prime number. Set $S=\mathbb Z\setminus q\mathbb Z$. Analogously $D_P$ is a localization of $S^{-1}D=\mathbb Z_{(q)}[\frac Xq]$ (here $\mathbb Z_{(q)}$ stands for the localization of $\mathbb Z$ at the prime ideal $q\mathbb Z$) which is also a noetherian ring.

Remark. The foregoing shows that $\dim D=2$.

All the previous answers send us to complicated examples since these were $1$-dimensional domains. But the OP has looked for

A domain $D$ all of whose localizations $D_P$ for $P∈\mathrm{Spec}(D)$ are noetherian, and $D$ is not noetherian.

A simple example is the following: $D=\mathbb Z[\frac Xp:p\text{ prime},p\ge 2].$
This is an integral domain which is not noetherian: $(\frac{X}{p_1})\subsetneq(\frac{X}{p_1p_2})\subsetneq\cdots$.
Now let $P$ be a prime ideal of $D$. There are two cases:
$\bullet$ $P\cap\mathbb Z=(0)$. Set $S=D\setminus\{0\}$. Then $D_P\simeq(S^{-1}D)_{S^{-1}P}$, that is, $D_P$ is a localization of $S^{-1}D=\mathbb Q[X]$ which is a noetherian ring.
$\bullet$ $P\cap\mathbb Z=q\mathbb Z$ with $q\ge2$ a prime number. Set $S=\mathbb Z\setminus q\mathbb Z$. Analogously $D_P$ is a localization of $S^{-1}D=\mathbb Z_{(q)}[\frac Xq]$ (here $\mathbb Z_{(q)}$ stands for the localization of $\mathbb Z$ at the prime ideal $q\mathbb Z$) which is also a noetherian ring.

Remark. The foregoing shows that $\dim D=2$.

All the previous answers send us to complicated examples since these are $1$-dimensional domains. But the OP has looked for

A domain $D$ all of whose localizations $D_P$ for $P∈\mathrm{Spec}(D)$ are noetherian, and $D$ is not noetherian.

A simple example is the following: $D=\mathbb Z[\frac Xp:p\text{ prime},p\ge 2].$
This is an integral domain which is not noetherian: $(\frac{X}{p_1})\subsetneq(\frac{X}{p_1p_2})\subsetneq\cdots$.
Now let $P$ be a prime ideal of $D$. There are two cases:
$\bullet$ $P\cap\mathbb Z=(0)$. Set $S=D\setminus\{0\}$. Then $D_P\simeq(S^{-1}D)_{S^{-1}P}$, that is, $D_P$ is a localization of $S^{-1}D=\mathbb Q[X]$ which is a noetherian ring.
$\bullet$ $P\cap\mathbb Z=q\mathbb Z$ with $q\ge2$ a prime number. Set $S=\mathbb Z\setminus q\mathbb Z$. Analogously $D_P$ is a localization of $S^{-1}D=\mathbb Z_{(q)}[\frac Xq]$ (here $\mathbb Z_{(q)}$ stands for the localization of $\mathbb Z$ at the prime ideal $q\mathbb Z$) which is also a noetherian ring.

Remark. The foregoing shows that $\dim D=2$.

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user26857
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