Flaws in my proof that a finite union of closed sets is closed (Proof Verification) (ps: I'm teaching myself analysis the first time as a sophomore in high school using Understanding Analysis, which means that I do not have so much experience in dealing with serious mathematics, so please be patient with me. Thanks!)
I'm trying to prove that a finite union of closed sets is also closed using the definition of limit point (that I summarized) as
$$
x\textrm{ is a limit point of } F \Leftrightarrow \forall \epsilon > 0, \exists a \in F, \textrm{such that } a \in V_{\epsilon}(x) \cap F \textrm{ and } a\neq x
$$
where $V_{\epsilon}(x)$ is the $\epsilon$-neighborhood of $x$.
Here is my attempt:

*

*Goal: prove that if $F_{i}$ closed $\forall 1\le i \le n$, then $\displaystyle \bigcup^{n}_{i=1}F_{i}$ is also closed.


*My proof: if $x$ is a limit point of $\bigcup^{n}_{i=1}F_{i}$, then $\forall \epsilon > 0, \exists a \in \displaystyle \bigcup^{n}_{i=1}F_{i},$ such that $a \in V_{\epsilon}(x) \cap \left(\displaystyle \bigcup^{n}_{i=1}F_{i}\right)$ and $a\neq x$. Now since $a \in V_{\epsilon}(x) \cap \left(\displaystyle \bigcup^{n}_{i=1}F_{i}\right)$, $a$ must be in $\bigcup^{n}_{i=1}F_{i}$, and therefore at least one $F_{i}$. This means that $\exists 1 \le i \le n, \forall \epsilon > 0, \exists a \in F_{i}$, such that $ a \in V_{\epsilon}(x) \cap F_{i} \textrm{ and } a\neq x$, hence $x$ is a limit point of $F_{i}$ for some $i$. Since each $F_{i}$ is closed (this is our assumption), $x \in F_{i}$ for some $i$, which guarantee that $x \in \bigcup^{n}_{i=1}F_{i}$. We picked an arbitrary $x$ in the beginning as a limit point of $\bigcup^{n}_{i=1}F_{i}$ and have just proven that $x$ is in $\bigcup^{n}_{i=1}F_{i}$, indicating that the union is a closed set.
Now here's my real question: I have (seemingly) proven the theorem I claimed I would without employing the fact that the union is finite, which means that (it seems like) my proof can be generalized to prove any arbitrary union of closed sets is also closed, which is certainly false. So where have I made a mistake in my proof? Is there a way to fix it without changing the climate of the entire proof?
(ps: I have seen people using the complement of open sets to prove this theorem, or other proofs using contradictions. I wholeheartedly accept both of those ways, but I just wanted to explore proving it directly.)
 A: It's better to assume that $x$, the limit point of $\bigcup_{i=1}^n F_i$ is not a limit point of any $F_i,i=1,2,\ldots,n$. That gives you $n$ many $\varepsilon_i>0$ that witness the "non-limit pointness" so $V_{\varepsilon_i}(a) \cap F_i \subseteq \{a\}$. (The intersection is empty or just $\{a\}$).
Then $\varepsilon = \min_{i=1}^n \varepsilon_i >0$ has $$V_\varepsilon(a) \cap \left( \bigcup_{i=1}^n F_i \right) \subseteq \{a\}$$ contradicting that $a$ is a limit point of the union.
Going from the contrapositive frees us from the dependence of $i$ on $\varepsilon$ that your proof attempt suffers from.
You show $$\forall \varepsilon>0: \exists i: V_\varepsilon(a)\cap F_i\setminus \{a\} \neq \emptyset$$ while you must show
$$\exists i: \forall \varepsilon>0: V_\varepsilon(a)\cap F_i\setminus \{a\} \neq \emptyset$$
(to show it is a limit point of some $F_i$) Order of quantifiers is important...
A: To understand where an argument goes wrong it's often useful to look at examples. So let's take an infinite collection of $F_i$ whose union is not closed: $F_i = \{1/i\}$ for each $i \ge 1$ as a subset of $\mathbb{R}$ with the usual metric. Then $0$ is a limit point of the union $\{1/i\ |\ i \ge 1\}$ but is not a limit point of any single $F_i$. Indeed, for any $\epsilon$ there is an $a \in \bigcup F_i$ and hence some $F_i$ such that $a$ is with $\epsilon$ of $0$ but this $i$ depends on $\epsilon$ and no fixed $i$ will work for every $\epsilon$.
Thinking another way, for fixed $\{F_i\}$ and $x$ let $S_i = \{\epsilon > 0 \ |\ F_i \cap V_\epsilon(x) \ne \emptyset \}$. Then we know that each $S_i$ is upward closed (if $\epsilon_1 \in S_i$ and $\epsilon_2 > \epsilon_1$ then $\epsilon_2 \in S_i$) and that $\bigcup S_i = (0, \infty)$ and from this we want to conclude that some $S_i = (0, \infty)$. Any "direct" proof of this fact (when there are only finitely many $i$) should translate to a fix for your proof:

*

*Using contrapositive (or contradiction) as in @HennoBrandsma's answer corresponds to the observation that if $S_i \ne (0, \infty)$ then $S_i \subset [c_i, \infty)$ for some $c_i>0$ but then $\bigcup S_i \subset [\min c_i, \infty)$. As a slight reformulation we could take $c_i = \inf S_i = d(x, F_i)$, but I don't think this avoids contrapositives.

*The argument in @DavidMitra's comment is akin to noticing that each $1/n$ is in some $S_i$ and therefore infinitely many $1/n$ must be in a fixed $S_i$.

