$A_i$ are closed. Any compact set intersects with finitely many $A_i$. Is $\bigcup_{i\in\mathbb N} A_i$ closed? I do not know if this proposition is true.
My guess: yes?
my attempts:
Take a convergent net $a_\lambda\to a_0$ from $\bigcup_{i\in\mathbb N} A_i$.

*

*Suppose there is some $i$ such that for every $\lambda$ there is $\eta\geq\lambda$ such that $a_\eta\in A_i$. Then $a_\eta$ such that $a_\eta\in A_i$ forms a subnet of $a$ and therefore converges to $a_0$. This shows $a_0\in A_i$.


*Suppose there is not, so that there are increasing $\lambda_i$ such that element of $A_i$ does not appear after $\lambda_i$. Samely, $a_{\lambda_i}\to a_0$, and therefore $\{a_0,a_{\lambda_1},...\}$ is compact. By the assumption it only intersects with finitely many $A_i$, but this is a contradiction.
I did it! Is this proof correct then?
By the way, is there some weaker condition which makes $\bigcup_{i\in\mathbb N} A_i$ a closed set?
 A: The result is false, I’m afraid.
Let $p$ be a free (i.e., non-principal) ultrafilter on $\Bbb N$, and let $X=\Bbb N\cup\{p\}$. Let
$$\tau=\wp(\Bbb N)\cup\big\{\{p\}\cup A:A\in p\big\}\,;$$
$\tau$ is a topology on $X$. For $n\in\Bbb N$ let $A_n=\{n\}$. The sets $A_n$ are closed, and the only compact sets in $X$ are the finite sets, so every compact subset of $X$ intersects only finitely many of the sets $A_n$. However, $\bigcup_{n\in\Bbb N}A_n=\Bbb N$, which is not closed: $p$ is a limit point of $\Bbb N$.
The problem with your argument is that the sequence $\langle a_{\lambda_i}:i\in\Bbb N\rangle$ need not converge to $a_0$.
Added: To see that every compact subset of $X$ is finite, let $A\subseteq X$ be infinite. Clearly $A$ is not compact if $A\subseteq\Bbb N$, so suppose that $p\in A$. If $A\notin p$, then $\Bbb N\setminus A\in p$, so $X\setminus A$ is an open nbhd of $p$, and
$$\{X\setminus A\}\cup\big\{\{n\}:n\in A\big\}$$
is an open cover of $A$ with no finite subcover. If $A\in p$, let $\{B,C\}$ be a partition of $A$ into two infinite sets. Since $p$ is an ultrafilter, exactly one of $B$ and $C$ belongs to $p$; without loss of generality suppose that $B\in p$. Then
$$\big\{\{p\}\cup B\big\}\cup\big\{\{n\}:n\in A\setminus B\big\}$$
is an open cover of $A$ with no finite subcover.
A: If $(A_i)_{i\in I}$ is a family of closed sets and is also locally finite ( every point $x\in X$ is contained in some $U_x$ open such that $U_x$ intersects finitely many of the $A_i$'s ) then $\cup_{i \in I} A_i$ is closed.
For proof: a set is closed if and only it is locally closed ( since it is true for "open").
Now, if your space is also locally compact then OK.
