I'm facing some difficulties into understanding the proof of the famous "Chevalley Theorem" in Milne's "Algebraic Geometry" (page 196)
A regular morphism $\varphi:W\to V$ between algebraic varieties sends constructible sets to constructible sets. In particular, if $U$ is a nonempty open subset of $W$, then $\varphi(U)$ contains a nonempty open subset of its closure in $V$.
EDIT: Thanks to KReiser and Aphelli, now my only doubt is the third and last one, so there's no need to address the first two doubts.
My attempt of grasping the proof
The author first proves the "In particular" part of the statement and says that
"by considering suitable open affine coverings of $W$ and $V$, we see that it suffices to prove this in the case that both $V$ and $W$ are affine varieties".
I think that I need more detail on this step. Let $\mathcal{U}=\{U_i\}$ be an affine covering of $W$. Clearly by choosing wisely the covering I can suppose that $f(U_i)$ is contained in an affine open set $\widetilde{U}_i$. Let's suppose that I already proved the "in particular" part of the statement for affine varieties so that $\varphi:U_i\to \widetilde{U}_i$ satisfies the "in particular" property. This means that $\varphi(U\cap U_i)$ contains an open subset $A_i$ of $\overline{\varphi(U\cap U_i)}\cap \widetilde{U}_i$ (I use the overline just for the closure in $V$ and in this case we are talking about the closure in $\widetilde{U}_i$). Clearly $A_i$ can be written as $\widetilde{A}_i\cap \overline{\varphi(U\cap U_i)}\cap \widetilde{U}_i$ where $\widetilde{A}_i$ is an open subset of $V$. So:
$$\varphi(U)=\bigcup_i \varphi(U\cap U_i)\supseteq \bigcup_i A_i \supseteq \left(\bigcup_i \overline{\varphi(U\cap U_i)}\right)\cap \bigcap_i \widetilde{A}_i\cap\widetilde{U}_i=\overline{\varphi(U)}\cap \bigcap_i \widetilde{A}_i\cap\widetilde{U}_i $$
and this would conclude the problem if I knew that $\bigcap_i \widetilde{A}_i \cap \widetilde{U}_i\neq \varnothing$. Am I overcomplicating this step? I don't see a simple way to do this.
Moving along with the proof, the author says that we can also suppose $W$ to be irreducible because if $W_1,...,W_r$ are its irreducible components, then $$\varphi(U)=\bigcup_j \varphi(U\cap W_j)$$ so I think that using a reasoning similar to the one of the previous step we can limit ourselves to the case $W$ irreducible (but I still don't know how to complete properly this reasoning).
Now the author says that we can also suppose $V$ to be irreducible and I'm not really convinced by the motivation:
"We can restrict the codomain $V$ to the closure of the image."
We are basically saying that if $\varphi(U)$ contains a nonempty subset of $\overline{\varphi(U)}\cap \overline{\varphi(W)} $ then it contains a nonempty subset of $\overline{\varphi(U)}$ and I don't understand why.
Moving along with the proof. Since the map $\varphi:W\to \overline{\varphi(W)}$ is dominant the image contains a dense open subset of the codomain, so the "in particular" part is proved.
Now we have to prove that th image of a constructible set is constructible. I understand that we can suppose $W$ and $C$ to be irreducible (clearly this implies that $\overline{C}$ is a closed subvariety of $W$). The theorem is now proven by induction on $\dim(W)$. The case $\dim(W)=0$ is trivial.
The inductive step is trivial if $\overline{C}\neq W$ because then $\dim(\overline{C})<\dim(W)$ (because of the irreducibility assumptions) and I can apply the inductive hyphothesis to $\varphi|_{\overline{C}}$ (clearly $C$ is also constructible in $\overline{C}$).
The hard case if $\overline{C}=W$. In this case $\overline{C}$ contains a nonempty open subset $O$ of $W$. Now the author comes to another tricky point
"The case just proved shows that $\varphi(C)$ contains a nonempty subset $U$ of its closure (in $V$)"
I've tried to understand this part with the following reasoning. By a precedent proposition I know that $C$ contains a nonempty open subset $\widetilde{O}$ of $\overline{C}$. Let $O\cap \widetilde{O}$. Clearly $\varphi(C)\supseteq \varphi(O\cap \widetilde{O})$ and by the "in particular" part of the proposition $\varphi(U)$ contains a nonempty open subset $U$ of the closure $\overline{\varphi(O\cap \widetilde{O})}$. So $\varphi(C)$ contains $U$ but this doesn't prove anything because I don't know if $U$ is open in $\overline{\varphi(C)}$.
Moving along with the proof, we can write
$$\varphi(C)=U\cup \varphi(C\cap \varphi^{-1}(\overline{\varphi(C)}-U))$$
Since $\varphi$ is continous $\varphi^{-1}(\overline{\varphi(C)}-U)$ is closed and it's a proper closed set because $\overline{C}=W$ and I can suppose $\varphi$ to be dominant (by simply restricting its codomain to the closure of the image). Clearly $C\cap \varphi^{-1}(\overline{\varphi(C)}-U)$ is constructible in $\varphi^{-1}(\overline{\varphi(C)}-U)$ so by inductive hyphothesis $\varphi(C\cap \varphi^{-1}(\overline{\varphi(C)}-U))$ is constructible in $V$ and this clearly implies that $\varphi(C)$ is constructible.
I'd be grateful if you could help me with the tricky steps. Even if you know how to help me just on some steps I'd be happy.
