Proposition II.$3.2$ in Hartshorne In Proposition II. (3.2) of Hartshorne book "algebraic geometry". I can't understand proof of part. 
Proposition:  A scheme $X$ is locally Noetherian iff for every open
affine $U = \rm Spec A$, $A$ is a Noetherian ring.
It suffices to prove that if $X$ is locally Noetherian, and $U =
\rm Spec A$ is an open affine subset then $A$ is a Noetherian ring.
We first show that $U$ is locally Noetherian. Suppose that $V =\rm Spec B$
is an open affine on $X$ where $B$ is a Noetherian ring. Then $U\cap V$ can
be covered by open sets of the form $D(f)\cong\rm Spec(B)$, where $f\in B.$ As $B$
is a Noetherian ring then so is $B_f.$ As open sets of the form $V$ cover
$X,$ $U$ is covered by open affines, which are the spectra of Noetherian
rings. So $U$ is locally Noetherian.
Replacing $X$ by $U,$  $^{(1)}$ we are reduced to proving that if $X =\rm Spec A$
locally Noetherian then $A$ is Noetherian.  Let $V =\rm Spec B,$ be an open
subset of $X$ , where $B$ is a Noetherian ring. Then there is an element
$f\in A$ such that $D(f)\subset V .$ Let $\bar f$ be the image of $f$ in $B.$          $^{(2)}$ Then $A_f\cong B_{\bar f}$   whence $A_f$ is Noetherian.
So we can cover $X$ by open subsets  $D(f)\cong \rm {Spec}$$(A_f).$ Since $X$ 
is quasi-compact, a finite number will do. 


*

*I don't see why we may replace $X$ by $U$ ? and how we reduce to prove that if $X =\rm Spec A$
locally Noetherian then $A$ is Noetherian.

*I don't understand what he mean by "image of $f$ in $B$" and why $A_f\cong B_{\bar f}.$


Thank you.

I would like to complement the answer to question 2 by expanding a little bit on Hoot's remark.
Let $X_f$ be the set of points of $X$ such that the germ of $f$ at each such point is not inside the maximal ideal of the corresponding local ring. If $U \cong \operatorname{Spec}B$ is an open affine set of $X$, then by exercise II.2.16(a) in Hartshorne, we have an isomorphism of schemes $X_f \cap U \cong D(\bar{f})$, where $\bar{f}$ is the image of $f$ under the ring homomorphism $A\rightarrow B$ that is associated to the inclusion $\operatorname{Spec}B \hookrightarrow \operatorname{Spec}A$. 
Now, if it so happens that $X_f$ is a subscheme of $U$, then we have an isomorphism of schemes $X_f \cong D(\bar{f})$. If in addition $X=\operatorname{Spec}A$, then we get $\operatorname{Spec}A_f \cong \operatorname{Spec}B_{\bar{f}}$. Since this is an isomorphism of schemes, this gives $A_f \cong B_{\bar{f}}$.
