We reduce the theorem of Chevalley into the proposition stated in the exercise by several steps.
To do so, we need some lemmas.
Notation
Let $X$ be a scheme.
We denote the ring of global sections of the structure sheaf $\mathcal{O}_X$ of $X$ by $\Gamma(X, \mathcal{O}_X)$.
By abuse of notatation, we often write $\Gamma(X)$ instead of $\Gamma(X, \mathcal{O}_X)$
if there is no risk of ambiguity.
Lemma 1
Let $f\colon X \rightarrow Y$ be a morphism of schemes.
Suppose $f$ is a morphism of finite type and $Y$ is a noetherian scheme.
Then $X$ is a noetherian scheme.
Proof:
Let $x$ be a point of $X$.
Since $f$ is of finite type, there exist an affine open neighborhood $U$ of $x$
and an affine open neighborhood $V$ of $f(x)$ such that $f(U) \subset V$
and the induced ring map $\Gamma(V) \rightarrow \Gamma(U)$ is of finite type.
Since $Y$ is a noetherian scheme, $\Gamma(V)$ is noetherian(Hartshorne II, Proposition 3.2).
Since $\Gamma(V) \rightarrow \Gamma(U)$ is of finite type, $\Gamma(U)$ is noetherian.
Hence $X$ is a locally noetherian scheme.
Since $f$ is quasi-compact and $Y$ is quasi-compact, $X$ is quasi-compact.
Hence $X$ is a noetherian scheme.
QED
Lemma 2
Let $X$ be a noetherian scheme.
Let $U$ be an open subset of $X$.
Then the canonical injection $f\colon U \rightarrow X$ is a morphism of finite type.
Proof:
It is clear that $f$ is locally of finite type.
Since $X$ is noetherian, every subset of $X$ is quasi-compact.
Hence $f$ is quasi-compact.
Therefore $f$ is of finite type.
QED
Lemma 3
Let $Z$ be a closed subscheme of a scheme $X$.
Then the canonical injection $f\colon Z \rightarrow X$ is a morphism of finite type.
Proof:
Let $V$ be an open affine subset $X$.
Let $A = \Gamma(V, \mathcal{O}_X)$.
Then the closed subscheme $Z \cap V$ of $V$ is isomorphic to Spec($A/I$)
for some ideal $I$ of $A$. Hence $f$ is locally of finite type.
Since Spec($A/I$) is quasi-compact, $f$ is quasi-compact.
Hence $f$ is of finite type.
QED
Lemma 4
Let $X$ be a noetherian scheme.
Let $Z$ be a subscheme of $X$.
Then the canonical injection $f\colon Z \rightarrow X$ is a morphism of finite type.
Proof:
There exists an open subscheme $U$ of $X$ such that $Z$ is a closed subscheme of $U$.
Let $h\colon Z \rightarrow U$ and $g\colon U \rightarrow X$ be the canonical injections.
By Lemma 2, $g$ is of finite type.
By Lemma 3, $h$ is of finite type.
Hence $f = g\circ h$ is of finite type.
QED
Lemma 5
Let $(X_i)$ be a finite family of schemes of finite type over a scheme $S$.
Then the coproduct $\sqcup X_i$ is of finite type over $S$.
Proof:
Clear.
Definition
Let $X$ be a topological space.
A finite union of locally closed subset of $X$ is called quasi-constructible.
If $X$ is a noetherian topological space, a quasi-constructible subset of $X$ is called
constructible.
Lemma 6
Let $X$ be a topological space.
Let $W$ be a locally closed subset of $X$.
Let $Z$ be a subset of $Z$.
Then $Z$ is a locally closed subset of $X$ if and only if it is a locally closed subset of $W$.
Proof:
Suppose $Z$ is a locally closed subset of $X$.
There exist an open subset $U$ of $X$ and a closed subset $F$ of $X$ such that $Z = U \cap F$.
Since $Z = Z \cap W = U \cap F \cap W$, $Z$ is a locally closed subset of $W$.
Conversely suppose $Z$ is a locally closed subset of $W$.
There exist an open subset $U$ of $X$ and a closed subset $F$ of $X$ such that $Z = U \cap F \cap W$. Hence $Z$ is a locally closed subset of $X$.
Lemma 7
Let $X$ be a topological space.
Let $W$ be a locally closed subset of $X$.
Let $Z$ be a subset of $W$.
Then $Z$ is a quasi-constructible subset of $X$ if and only if it is a quasi-constructible subset of $W$.
Proof:
This follows immediately from Lemma 6.
Lemma 8
Let $X$ be a topological space.
Let $(W_i)$ be a finite cover of $X$.
Suppose each $W_i$ is a locally closed subset of $X$.
Let $Z$ a subset of $X$.
Then $Z$ is a quasi-constructible subset of $X$ if and only $Z \cap W_i$ is a quasi-constructible subset of $W_i$ for every $i$.
Proof:
This follows immediately from $Z = \bigcup_i (Z \cap W_i)$ and Lemma 7.
Lemma 9
Let $f\colon X \rightarrow Y$ be a morphism of affine schemes.
Then there exist a closed subscheme $Z$ of $Y$ and a morphism $g\colon X \rightarrow Z$
such that $f = j\circ g$ and $g(X)$ is dense in $Z$,
where $j\colon Z \rightarrow X$ is the canonical morphism.
Moreover, if $f$ is of finite type, $g$ is of finite type.
Proof:
Suppose $X =$ Spec$(B)$, $Y =$ Spec$(A)$.
Let $\psi\colon A \rightarrow B$ be a homomoprphism which induces $f$.
Let $I$ be the kernel of $\psi$.
Then $C = \psi(A)$ is canonically isomorphic to $A/I$.
$\psi$ factors into $A \rightarrow C \rightarrow B$, where $C \rightarrow B$ is the canonical injection.
If $B$ is of finite type over $A$, $B$ is of finitye over $C$.
Letting $Z =$ Spec$(C)$ we are done.
QED
We will reduce the theorem in several steps.
Step 1.
We may assume $Z = X$.
Proof:
Let $Z = (U_1 \cap F_1)\cup \cdots \cup (U_n \cap F_n)$, where $U_i$ is an open subset of $X$ and $F_i$ is a closed subset of $X$.
Let $Z_i = U_i\cap F_i$ for each $i$.
We consider each $Z_i$ as a reduced closed subscheme of $U_i$.
By Lemma 4, the canonical morphism $Z_i \rightarrow X$ is of finite type.
Let $Z' = \sqcup Z_i$ be a coproduct of schemes.
Let $g\colon Z' \rightarrow X$ be the canonical morphism.
By Lemma 5, $g$ is of finite type.
Since $g(Z') = Z$, $fg(Z') = f(Z)$.
Since $fg$ is of finite type, we may replace $f$ with $fg$.
Step 2.
We may assume $Y$ is affine.
Proof:
Let $(V_i)$ be a finite open affine cover of $Y$.
By Lemma 8, $f(X)$ is constructible in $Y$ if $f(X) \cap V_i$ is constructible in $V_i$ for all $i$.
Let $f_i\colon f^{-1}(V_i) \rightarrow V_i$ be the restriction of $f$.
Then $f_i(f^{-1}(V_i)) = f(X) \cap V_i$.
Clearly $f_i$ is of finite type.
Hence it suffices to prove that $f_i(f^{-1}(V_i))$ is constructible in $V_i$ for each i.
Step 3.
We may assume $X$ is affine.
Proof:
Let $(U_i)$ be a finite open affine cover of $X$.
Let $g_i\colon U_i \rightarrow X$ be the canonical ingection for each $i$.
Let $f_i = f\circ g_i$.
Since $X$ is a noetherian scheme by Lemma 1, $g_i$ is of finite type by Lemma 2.
Since $f$ is of finite type, $f_i$ is of finite type.
Since $f(X) = \bigcup f_i(U_i)$, it suffices to prove $f_i(U_i)$ is constructible in $Y$ for each $i$.
Step 4.
We may assume $Y$ is irreducible.
Proof:
Let $Y_1,\dots,Y_m$ be irreducible components of $Y$.
By Lemma 8, it suffices to prove $f(X) \cap Y_i$ is constructible in $Y_i$ for each $i$.
We regard $Y_i$ as a reduced closed subscheme of $Y$.
Let $f_i\colon X\times_Y Y_i \rightarrow Y_i$ be the canonical morphism.
Since $f$ is of finite type, $f_i$ is of finite type.
Since $f(X) \cap Y_i = f_i(f^{-1}(Y_i))$ and $f^{-1}(Y_i) = X\times_Y Y_i$, we may replace $f$ with $f_i$.
Step 5.
We may assume $X$ is irreducible.
Proof:
Let $X_1,\dots,X_n$ be irreducible components of $X$.
Since $f(X) = \bigcup f(X_i)$, it suffices to prove $f(X_i)$ is constructible in $Y$ for each $i$.
We regard each $X_i$ as a reduced closed subscheme of $X$.
Let $g_i\colon X_i \rightarrow X$ be the canonical morphism.
Since $g_i$ is of finite type by Lemma 3, $fg_i$ is of finite type.
Hence we may replace $f$ with $f_i = fg_i$.
Step 6.
We may assume $f$ is dominant.
Proof:
By Lemma 9, there exist a closed subscheme $Z$ of $Y$ and a morphism $g\colon X \rightarrow Z$
of finite type such that $f = j\circ g$ and $g(X)$ is dense in $Z$,
where $j\colon Z \rightarrow X$ is the canonical morphism.
By Lemma 7, if $f(X)$ is constructible in $Z$, $f(X)$ is constructible in $Y$.
Hence we may replace $f$ with $g$.
Step 7.
We may assume $X$ and $Y$ are integral.
Proof:
We may replace $f$ with $f_{red}\colon X_{red} \rightarrow Y_{red}$.
