Prove that a finite union of closed sets is also closed Let $X$ be a metric space.  If $F_i \subset X$ is closed for $1 \leq i \leq n$, prove that $\bigcup_{i=1}^n F_i$ is also closed.
I'm looking for a direct proof of this theorem.  (I already know a proof which first shows that a finite intersection of open sets is also open, and then applies De Morgan's law and the theorem "the complement of an open set is closed.")  Note that the theorem is not necessarily true for an infinite collection of closed $\{F_\alpha\}$.
Here are the definitions I'm using:
Let $X$ be a metric space with distance function $d(p, q)$.  For any $p \in X$, the neighborhood $N_r(p)$ is the set $\{x \in X \,|\, d(p, x) < r\}$.  Any $p \in X$ is a limit point of $E$ if $\forall r > 0$, $N_r(p) \cap E \neq \{p\}$ and $\neq \emptyset$.  Any subset $E$ of $X$ is closed if it contains all of its limit points.
 A: Here is one method, that I think is very direct:
Check first that a set contains all limit points if and only if every converging sequence in the set has a limit in the set. Now take a convergent sequence in the finite union. Since the union is finite, one of the sets in the union must contain infinitely many terms of the sequence and therefore a subsequence. A subsequence of a convergent sequence is converging and converges to the same point. So there is a converging subsequence lying whole in one of the sets of the finite union and this set contains the limit since it is closed. So the limit lies in the finite union, and we are done.

Edit:
Here is a sequence free version. Suppose $F_1$ and $F_2$ are closed. Let $x$ be a limit point of $F_1\cup F2$. We are done if we can show that $x$ is a limit point of $F_1$ or $F_2$. If $x$ is not a limit point of $F_1$, then there is an $\epsilon>0$ such that the $\epsilon$-Ball around $x$ contains no element of $F_1$. Hence it contains a point from $F_2$ and by the definition of a limit point, for every positive $\epsilon'<\epsilon$, the $\epsilon'$-Ball contains an element of $F_2$. Hence, $x$ is a limit point of $F_2$.
A: It is sufficient to prove this for a pair of closed sets $F_1$ and $F_2$. Suppose $F_1 \cup F_2$ is not closed, even though $F_1$ and $F_2$ are closed. This means that some limit point $p$ of $F_1 \cup F_2$ is missing. So there is a sequence $\{ p_i\} \subset F_1 \cup F_2$ converging to $p$. By pigeonhole principle, at least one of $F_1$ or $F_2$, say $F_1$, contains infinitely many points of $\{p_i\}$, hence contains a subsequence of $\{p_i\}$. But this subsequence must converge to the same limit, so $p \in F_1$, because $F_1$ is closed. Thus, $p \in F_1 \subset F_1 \cup F_2$.
Alternatively, if you do not wish to use sequences, then something like this should work. Again, it is sufficient to prove it for a pair of closed sets $F_1$ and $F_2$. Suppose $F_1 \cup F_2$ is not closed. That means that there is some points $p \notin F_1 \cup F_2$ every neighbourhood of which contains infinitely many points of $F_1 \cup F_2$. By pigeonhole principle again, every such neighbourhood contains infinitely many points of at least one of $F_1$ or $F_2$, say $F_1$. Then $p$ must be a limit point of $F_1$; so $p \in F_1 \subset F_1 \cup F_2$.
A: Let $F$ and $G$ be two closed sets and let $x$ be a limit point of $F\cup G$. Now, if $x$ is a limit point of $F$ or $G$ it is clearly contained in $F\cup G$. So suppose that $x$ is not a limit point of $F$ and $G$ both. So there are radii $\alpha$ and $\beta$ such that $N_\alpha(x)$ and $N_\beta(x)$ don't intersect with $F$ and $G$ respectively except possibly for $x$. But then if $r=min (\alpha,\beta)$ then $N_r(x)$ doesn't intersect with $F\cup G$ except possibly for $x$, which contradicts $x$ being a limit point. This contradiction establishes the result. The proof can be extended easily to finitely many closed sets. Trying to extend it to infinitely many is not possible as then the "min" will be replaced by "inf" which is not necessarily positive. 
