Is the restriction of a finite map of affine varieties also finite? If $f:X\to Y$ is a dominant(i.e.$f(X)$ is dense in $Y$) regular map of affine varieties, then $f$ is called a finite map if $k[X]$ is integral over $f^*k[Y]$.

My question is: if $Z\subset X$ is a closed subset of $X$, then how to
  show the restiction $f|_Z: Z\to \overline{f(Z)}$ is still a finite
  map?

(Note: This result is supposed to prove the fact that "a finite map takes closed sets to closed sets"(cf.Page60, Basic Algebraic Geometry 1, by Shafarevich), so please do not use the latter fact in the answer.)
 A: Regular map  $f:X\to Y$ of affine varieties correspond bijectively  to morphisms of $k$-algebras $\phi: k[Y]\to k[X]$.
The map $f$ is dominant iff $\phi$ is injective,  and is (by definition) finite  iff $\phi$ makes $k[X]$ a finite $k[Y]$- module   i.e.  if $k[X]$ is a module of finite type over $\phi(k[Y])$ .
[The definition in Shafarevich is equivalent but confuses the issue with irrelevant integrality conditions.]
For any subset $Z\subset X$ the induced morphism $\bar \phi:k[Y]/\phi ^{-1}(I(Z))\to k[X]/I(Z) $ is also injective and module-finite so that the corresponding geometric restriction map map $\operatorname {res}(f): Z\to \overline  {f(Z)}$ is  dominant and finite, just as $f$ was.  
NB
Notice that, in conformity with your request, I have never mentioned closedness of any map.
A: A dominant map of affine varieties corresponds to an inclusion of rings $R \subseteq S$, where $S$ is integral over $R$.  You can identify $X, Y$ with the maximal spectra of $S, R$, and $f: X \rightarrow Y$ is the contraction map $\mathfrak n \mapsto \mathfrak n \cap R$.
A closed subset $Z$ of $X$ corresponds to a radical ideal $J$ of $S$.  Specifically, $J$ is the intersection of all the maximal ideals in $Z$.  Finite morphisms of varieties are closed, so $f(Z)$ corresponds to a radical ideal $I$ of $R$.  One can identify $Z$ with the maximal spectrum of $S/J$, and $f(Z)$ with the maximal spectrum of $R/I$.  
Concretely, $f(Z)$ is the set of all $\mathfrak n \cap R : \mathfrak n \in Z$, and so $I$ is equal to $$\bigcap\limits_{\mathfrak m \in f(Z)} \mathfrak m = \bigcap\limits_{\mathfrak n \in Z} (\mathfrak n \cap R) = R \cap \bigcap\limits_{\mathfrak n \in Z}  \mathfrak n = R \cap J$$
The composition $Z \rightarrow X \rightarrow f(Z)$ then corresponds to the ring homomorphism $\pi: R/I \rightarrow S/J$.  This is well defined and injective.  Since $S$ is integral over $R$, $S/J$ is integral over the image of $R/I$, hence $Z \rightarrow f(Z)$ is a finite, dominant morphism of varieties.
