mjhjmtu

Is a scheme being Noetherian equivalent to the underlying topological space being Noetherian and all its stalks being Noetherian?

This is false. The easiest counterexample I could come up with is the following "affine line with embedded points at every closed [rational] point":

Example. Let $k$ be an infinite field, let $R = k[x]$, and for each $alpha in k$ let $R_alpha = R[y_alpha]/((x-alpha)y_alpha,y_alpha^2)$. Then $R_alpha$ is an affine line with an embedded prime $mathfrak p_alpha = (x-alpha,y_alpha)$ at $x = alpha$, sticking out in the $y_alpha$-direction. Finally, let
$$R_infty = bigotimes_alpha in k R_alpha = operatorname*colim_substacklongrightarrow\I subseteq k\textfinite bigotimes_alpha in I R_alpha$$
be their tensor product over $R$ (not over $k$); that is
$$R_infty = frack[x]left[y_alpha_alpha in kright]sum_alpha in k((x-alpha)y_alpha, y_alpha^2).$$
This is not a Noetherian ring, because the radical $mathfrak r = (y_alpha_alpha in k)$ is not finitely generated. But $operatornameSpec R_infty$ agrees as a topological space with $operatornameSpec R_infty^operatornamered = mathbb A^1_k$, hence $|!operatornameSpec R_infty|$ is a Noetherian topological space.

On the other hand, the map $R to R_alpha$ is an isomorphism away from $alpha$, and similarly $R_alpha to R_infty$ induces isomorphisms on the stalks at $alpha$. Thus, the stalk $(R_infty)_mathfrak q_alpha = (R_alpha)_mathfrak p_alpha$ at $mathfrak q_alpha = mathfrak p_alpha R_infty + mathfrak r$ is Noetherian. Similarly, the stalk at the generic point $mathfrak r$ is just $R_(0) = k(x)$. Thus, we conclude that all the stalks of $R_infty$ are Noetherian. $square$

Remark. As requested in the comments, my motivation to come up with this example is the following: I was trying to prove by hand that the answer was positive. After an immediate reduction to the affine case, one needs to consider a chain $I_0 subseteq I_1 ldots$ of ideals. Because $|X|$ is Noetherian, we may assume they [eventually] define the same closed set $V$. Intuitively, to check that they agree as ideals, it should suffice to check it at each component of $V$. Then you can try to use the Noetherian rings $mathcal O_X,x$ for the generic points $x$ of $V$.

However this is not quite true, because of embedded points. The precise statement [Tags 0311 and 02M3] is that the inclusion $I_i-1 subseteq I_i$ is an equality if and only if the same holds for the localisation at every associated point of $I_i$. Now you run into trouble if the $I_i$ differ at an embedded point that is not seen by the topology of $V$. Moreover, if this embedded point varies as we move through the chain, there is a chance that the localisations $mathcal O_X,x$ at these embedded points are still Noetherian. Once you see this, coming up with the example is not so hard.

A further counterexample is Example 2.3 in W. Heinzer, J. Ohm, Locally noetherian commutative rings, Trans. Amer. Math. Soc. 158 (1971), 273-284.