Theorem   The union of a locally finite collection of closed subsets of a topological space is itself closed.

Proof. Let $\cal S$ be a locally finite collection of closed subsets of a topological space $X$ and put $Y=\cup\cal S$ Let $x\in X\setminus Y$ By local finiteness there is an open neighbourhood $U$ of $x$ that meets only finitely many members of $\cal S$ say $A_1,\dots,A_n$ So $U\setminus Y=U\setminus\bigcup_{i=1}^n A_i$ which is open. Thus $U\setminus Y$ is an open neighbourhood of $x$ that does not meet $Y$ It follows that $Y$ is closed.