A criterion for a finite union of locally closed subsets of an affine space to be closed in Zariski topology Let $k$ be a field.
Let $n > 0$ be an integer.
Let $K$ be an algebraically closed extension of $k$.
Suppose the trancendence dimension of $K/k \ge n$.
Let $A = k[x_1,\dots,x_n]$ be a polynomial ring.
Let $E$ be a subset of $A$.
We denote by $V(E)$ the common zeros of $E$ in $K^n$.
It is easy to see that by defining subsets of the form $V(E)$ as closed sets, we can define topology in $K^n$.
Let $C$ be a finite union of locally closed subsets of $K^n$.
Suppose the closure of $\{x\}$ is contained in $C$ for every $x \in C$.
Is $C$ closed?
 A: The answer is yes, and follows from the fact that a constructible set in a Noetherian scheme is closed if and only if it is closed under specialization.  
A: 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.
