Let $\{A_i\}_{i \in I}$ be a locally finite collection of closed subsets of $X$. Show that the union $\bigcup_{i \in I} A_i$ is closed in $X$. 
Let $\{A_i\}_{i \in I}$ be a locally finite collection of closed subsets of a topological space $X$. Show that the union $\bigcup_{i \in I} A_i$ is closed in $X$.

We are to show that $\left(\bigcup_{i \in I} A_i \right)^c = \bigcap_{i \in I}A_i^c$ is open. So let $x \in \bigcap_{i \in I}A_i^c$. Since $\{A_i\}_{i \in I}$ is locally finite there exists neighbourhood of $x$ say $O$ such that $O \cap A_n \ne \emptyset$ for $n \in \Bbb N$.
What results do I have to use now? I'm trying to show that we can find an open set containing $x$ that's a contained in $\bigcap_{i \in I}A_i^c$.

Suppose $x \notin \bigcup_{i \in I} A_i$, so $x$ is in none of the $A_i$.
The family being locally finite means there is some open neighbourhood $U_x$ of $x$ so that $F:=\{i \in I\mid A_i \cap U_x\neq \emptyset\}$ is a finite set of indices.
For each $i \in F$ we have $x \notin A_i$ so some $O_i$ open around $x$ exists such that $O_i \cap A_i =\emptyset$.
Now note that $O:=U_x \cap \bigcap_i O_i$ is an open subset containing $x$ (as all sets are open and $F$ is finite!) so that $$O \cap \left(\bigcup_{i \in I} A_i\right) = \emptyset$$ showing that the union is indeed closed.
($y \in O$ means $y \in U_x$ so that $y \notin A_i$ for $i \notin F$ already. For $ i \in F$ we use $y \in O_i$ so $y \notin A_i$ too).
