The following is a detailed proof of the fact stated in the answer by Matt E.
Notation Let $X$ be a topological space. Let $E$ be a subset of $X$. We denote by $cl(E)$ the closure of $E$ in $X$.
Definition 1 Let $X$ be a topological space. A finite union of locally closed subsets of $X$ is called a constructible subset of $X$.
Definition 2 Let $X$ be a topological space. Let $x \in X$. If $cl(\{x\}) = X$, $x$ is called a generic point of $X$.
Lemma 1(A slight generalization of Hartshorne, Ex. 3.18 (b), II) Let $X$ be an irreducible topological space. Suppose $X$ has a (not necessarily unique) generic point $\eta$. Let $C$ be a constructible subset of $X$. Suppose $C$ is dense in $X$. Then $C$ contains $\eta$. Moreover $C$ contains non-empty open subset of $X$.
Proof: Let $C = \bigcup_i (U_i \cap F_i)$ be a finite union of locally closed subsets of $X$, where $U_i$ is open and $F_i$ is closed. Since $cl (C) = \bigcup_i cl(U_i \cap F_i)$, $\eta \in cl(U_i \cap F_i)$ for some $i$. Since $cl(\{\eta\}) \subset cl (U_i \cap F_i)$, $X = cl(U_i \cap F_i)$. Since $\eta \in U_i$, $\eta$ is a generic point of $U_i$. Since $cl(U_i ∩ F_i) \cap U_i = U_i$, $U_i \cap F_i$ is dense in $U_i$. Since $U_i \cap F_i$ is a closed subset of $U_i$, $U_i \cap F_i = U_i$. Hence $\eta \in U_i \cap F_i \in C$. Since $U_i \cap F_i = U_i$, $U_i \subset C$. QED
Lemma 2(A slight generalization of Hartshorne, Ex. 3.18 (c), II) Let $X$ be a Noetherian topological space. Suppose every irreducible closed subset of $X$ has a generic point. Let $C$ be a constructible subsets of $X$. Suppose $cl(\{x\}) \subset C$ for every $x \in C$. Then $C$ is closed.
Proof: Since $X$ is Noetherian, the subspace $C$ is also Noetherian. Hence $C = C_1 \cup \cdots \cup C_n$, where each $C_i$ is closed and irreducible in the subspace $C$. Since $C_i = C \cap F_i$ for some closed subset $F_i$ of $X$, $C_i$ is constructible. Since $cl(C) = cl(C_1) \cup \cdots\cup cl(C_n)$ and each $cl(C_i)$ is irreducible, $C_i$ contains a generic point of $cl(C_i)$ by Lemma 1. Hence, by the assumption, $cl(C_i)$ is contained in $C$. Hence $cl(C) = C$. QED
Lemma 3 Let $K/k$ be as in the question. Then $X = K^n$ is Noetherian with the topology defined in the question and every irreducible closed subset of $X$ has a generic point.
Proof: Since $A = k[x_1,\dots, x_n]$ is a Noetherian ring, it is easy to see that $X$ is Noetherian. Let $Y$ be an irreducible closed subset of $X$. Then $Y = V(P)$, where $P$ is a prime ideal of $A$. Let $L$ be the field of fractions of $A/P$. Since the trancendence dimension of $L/k \le n$, there exists a $k$-homomorphism $\psi\colon L \rightarrow K$. Let $\bar x_i$ be the image of $x_i$ by the canonical homomorphism $A \rightarrow A/P$ for each $i$. Let $\psi(\bar x_i) = y_i$. Let $y = (y_1, \dots, y_n)$. Then $\psi$ induces a $k$-isomorphism $A/P \rightarrow k[y_1,\dots,y_n]$. Hence $f(y) = 0$ if and only if $f \in P$, where $f \in A$. Hence $y$ is a generic point of $Y$. QED
Proof of the assertion in the question This is immediate from Lemma 2 and Lemma 3.