morphism of schemes that is closed at topological space level

Dear saurav, yes the restriction is closed and you don't have at all to assume $k$ algebraically closed.

Reminder: For a scheme $S$ of finite type over $k$, the set $T_0$ of closed points of $T$ is very dense in $T$, i.e. dense in every closed subset of $T$ .

Here then is the statement you need :

Let $f:Xto Y$ be a closed morphism between schemes of finite type over a field $k$. Then the restriction $f_0:X_0to Y_0$ to the subspaces of respective closed points is closed.

Proof:

1) We may assume $X$ and $Y$ reduced. We have to show that, for $F$ closed in $X$, the subset $f(Fcap X_0)$ is closed in $Y_0$. By endowing $F$ and $f(F)$ with their reduced scheme structure we can assume $F=X, f(F)=Y$ and we have reduced (!) the problem to showing that $Y_0=f(X_0)$ : call this "closed surjectivity".

2) Here is why "closed surjectivity" is true. Take any closed point $y_0in Y_0$. It has a residual field $kappa (y_0)$ which is a finite extension of $k$.The fibre $Z=f^{-1}(y_0)$ is a closed non-empty subscheme of $X$ (recall that after our reduction $f$ is assumed surjective!).
But $Z$ is also a $kappa (y_0)$-scheme of finite type. Hence it has a closed point $x_0$, since those closed points are even very dense in $Z$. Since $Z$ is closed in $X$, $x_0$ is closed also in $X$ i.e. $x_0in X_0$. So we have shown "closed surjectivity" and everything is proved.

(Everything I used is contained in EGA I)