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.

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 $p \in F_1 \subset F_1 \cup F_2$.
