Application of ZMT to birational maps in Milne's Algebraic Geometry

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I'm studying Algebraic Geometry from this book. At page 187 the author proves that

If $\varphi:W\to V$ is a regular birational quasi-finite map between irreducible varieties and $V$ is normal, then $\varphi$ is an open immersion.

I understand many bits of this proof but I fail to grasp some fundamental steps.

My attempt of grasping the proof

Since $\varphi$ is a quasi-finite regular map, we can factor it thanks to ZMT (Zariski's Main Theorem)

$$W\overset{j}\hookrightarrow V'\overset{\varphi'}\to V $$

Since a finite morphism is an affine morphism I can cover $V$ with open affine sets $U$ so that $\varphi':\varphi'^{-1}(U)\to U$ is a finite morphism. This means that the extension $k[U]\hookrightarrow k[\varphi'^{-1}(U)]$ is integral. Now comes the first unclear step:

This means that $k[\varphi'^{-1}(U)]$ is the integral closure of $k[U]$ in $k(W)$.

Just... why? Why an integral extension is automatically the integral closure?

Moving along with the proof $k[\varphi'^{-1}(U)]$ is the integral closure of $k[U]$ in $k(W)$ and $k(W)=k(V)$ because $\varphi$ is birational. Now another step that I don't understand

Since $V$ is normal $k[U]$ is integrally closed in $k(V)$.

Every affine open set of a normal variety is normal so I agree that $k[U]$ is integrally closed in $k(U)$. But why in $k(V)$?

Moving along with the proof, since $k[U]$ is integrally closed in $k(V)$ and $k[\varphi'^{-1}(U)]$ is the integral closure of $k[U]$ in $k(V)$ we have that $k[U]=k[\varphi'^{-1}(U)]$ so $U$ is isomorphic to $\varphi'^{-1}(U)$. Now comes the last step that I don't understand.

This proves that $V$ is isomorphic to $V'$ and this concludes the proof.

I think that we only proved that $V$ and $V'$ are "locally isomorphic". And even assuming that $V\cong V'$, how does this conclude the proof exactly?

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I’m turning my comment into an answer.

For your first question, you are right: there is a priori no reason why $k[\varphi’^{-1}(U)]$ should be the integral closure of $k[U]$ in $k(W)$. The answer lies in the last sentence of Theorem 8.45, which gives an explicit factorization in this setting.

For your second question, note that $k(U)=k(V)$.

For your third question, I am less certain, because I didn’t read the entire proof. What seems most likely is that you don’t actually prove that $V’$ and $V$ are locally isomorphic (which wouldn’t imply anything, of course), but that $\varphi’: V’ \rightarrow V$ is an isomorphism locally on the base.

If this is the case, it then follows that $\varphi’$ is an actual isomorphism and thus that $\varphi=\varphi’ \circ j$ is an open immersion.