There was a counterexample given in the lecture which I didn't understand.
Let $A = \prod \limits_{i= 0}^{\infty} \mathbb{F}_2$. Then this is clearly not Noetherian, since we can take as a chain of non-stabilizing ideals $I_n:=\mathbb{F}_2 \times...\times \mathbb{F}_2 \times \{0\} \times \{0\}...$ where we have n-factors of $\mathbb{F}_2$.
But every localization $A_p$ at a prime ideal is a local ring and isomorphic to $\mathbb{F}_2$.
Now I don't understand the last part. I know that prime ideals of $A$ are a product of prime ideals of $\mathbb{F}_2$ and prime ideals $p$ of $\mathbb{F}_2$, so $(0)$.
Note that every element $x\in A$ satisfies $x^2=x$. Clearly, if $P\subseteq A$ is a prime ideal then the localization $A_P$ has the same property. So take any $x\in A_P$. Then $x^2=x$, and so $x(1-x)=0$. Since $A_P$ is a local ring, one of the elements $x$ and $1-x$ must be invertible (if both were non invertible, then $1=x+(1-x)$ would be non invertible as well, a contradiction), and so $x=0$ or $x=1$.
So it follows that $A_P$ has only two elements, $0$ and $1$.