Prove $\overline{A \cup B} = \overline A \cup \overline B$ Prove $\overline{A \cup B} = \overline A \cup \overline B$ (where $A,B$ are arbitrary sets, and $\overline S$ means the smallest closed set containing $S$).  Proofs for this are readily available; please verify and critique my proof.
Proof: We first prove that a limit point $\ell$ of $A \cup B$ must either be a limit point of $A$ or a limit point of $B$.
For any $\epsilon > 0$, there exists an open disc $D_\epsilon(\ell)$ with at least one point in $A \cup B$.  Suppose there exists $k > 0$ where there does not exist an open disc $D_k(\ell)$ with at least one point in $A$.  Then, for any $\epsilon \leq k$, there exists an open disc $D_\epsilon(\ell)$ with at least one point in $B$, and $\ell$ is therefore a limit point of $B$.  Conversely, if for all $k > 0$ there exists an open disc $D_k(\ell)$ with at least one point in $A$, $\ell$ is a limit point of $A$.
The result then follows from the definition of union, closure, and limit point, and the fact that the closure of a set is equal to the union of the set and its limit points.*
By induction, this can be extended to any finite collection of sets.  However, it cannot be extended to an infinite collection of sets, as evidenced by $0 \in \overline{\{1/1\}\cup  \{1/2\} \cup \{1/3\}...}$ but not in any $\overline{\{1/n\}}$.

*Update: The comments requested that I write out the part I asserted was trivial.  I still maintain that it is indeed trivial and should not be written.  Do you agree? If $\ell \in \overline{A \cup B}$, then either $\ell \in A \cup B$ or $\ell \in (A \cup B)'$ (with $S'$ meaning the limit points of $S$).  In the first case, $\ell$ is trivially in $\overline A \cup \overline B$, as any set is a subset of its closure.  The second case is addressed by the above. Conversely, if $\ell \in \overline A \cup \overline B$, then either $\ell \in A \cup B$, in which case $\ell \in \overline{A \cup B}$, or $\ell \in A' \cup B'$.  Since $S \subset U \implies S' \subset U'$, the latter case implies $\ell \in \overline{A \cup B}$ as well.
 A: Here are some critiques:
1.) I think the organization of the proof could be improved. For simple problems where you are showing two sets are equal, I generally think it's best to just show the two sets are subsets of each other. Your first paragraph sort of shows that $\overline{A \cup B} \subseteq \overline{A} \cup \overline{B}$, but it's good to make these things very clear (especially at the level of math where you're being asked this as an exercise). You also did not show the reverse containment.
2.) You end with "The result then follows from the definition of union, closure, and limit point, and the fact that the closure of a set is equal to the union of the set and its limit points.*" While you remarked why this is "trivial", if you write the proof in a more structured way you don't need to appeal to all these things to conclude your argument. From the view of a reader, when I read "the result follows by XYZ" and the XYZ is a long list of definitions, observations, etc. it would take me a minute to figure out how to put these together (worse yet, that I may lack trust you know how these things should fit together).
3.) This is a matter of notation/style, but usually $k$ (when meant to refer to a number) is reserved for integers, whereas you use it to mean a real number. In theory it doesn't matter, but sticking with conventional notation often improves clarity.
4.) You give the definition of $\overline{S}$ to be the smallest closed set containing $S$, but then use the fact that $x \in \overline{S}$ means $x \in S$ or $x$ is a limit point of $S$. Of course, this fact is extremely well-known, but if you are just working with the original definition you should at least mention that we can easily show the second fact is an equivalent definition of closure.
EDIT: As requested, how I would write the proof:
We'll do the proof assuming $A$ and $B$ are subsets of metric space $X$. For $\varepsilon > 0$ and $x \in X$, we denote $B_\varepsilon(x)$ to be the open ball of radius $\varepsilon$ centered at $x$. We'll use the stated fact that $\overline{S} = S \cup S'$, where $S'$ is the set of limit points of $S$, for any $S \subseteq X$.
Proof: Let $p \in \overline{A \cup B}$, so either $p \in A \cup B$ or $p \in (A \cup B)'$. In the former case, $p \in \overline{A} \cup \overline{B}$, since any set is a subset of its closure. In the latter case, let $\varepsilon > 0$. Since $p \in (A \cup B)'$, we know $B_\varepsilon(p) \cap (A\cup B)$ contains a point, $q$, distinct from $p$. We have:
$$
B_\varepsilon(p) \cap (A \cup B) = (B_\varepsilon(p) \cap A) \cup (B_\varepsilon(p) \cap B),
$$
so either $q \in B_\varepsilon(p) \cap A$ or $q \in B_\varepsilon(p) \cap B$. In other words, $p$ is either a limit point of $A$ or a limit point of $B$. This establishes $\overline{A\cup B} \subseteq \overline{A} \cup \overline{B}$.
Conversely, let $p \in \overline{A} \cup \overline{B}$, so $p \in \overline{A}$ or $p \in \overline{B}$. Without loss of generality, assume $p \in \overline{A}$. If $p \in A$, then $p \in \overline{A \cup B}$ immediately, so suppose $p \in A'$. Let $\varepsilon > 0$, and consider $B_\varepsilon(p) \cap (A \cup B)$. Since $p \in A'$, there exists $q \in B_\varepsilon(p) \cap A$ distinct from $p$. But then $q \in B_\varepsilon(p) \cap (A \cup B)$ as well, so $p \in (A\cup B)'$, and consequently $p \in \overline{A\cup B}$. This establishes $\overline{A} \cap \overline{B} \subseteq \overline{A \cap B}$. QED
A: Drawing on ccroth's proof and recommendations, while addressing the concerns I raised (in the comments), I present the revised approach below.  Please critique.
Lemma: $(A \cup B)' = A' \cup B'.$
Proof: For any $p \in (A \cup B)'$, we have that for any $\varepsilon > 0$, there exists a point $q \neq p$ such that $q \in B_\varepsilon(p) \cap (A \cup B)$, implying that $q \in B_\varepsilon(p) \cap A$ or $q \in B_\varepsilon(p) \cap B$.  Conversely, for any $p \in A'$, we have that for any $\varepsilon > 0$, there exists a point $q \neq p$ such that $q \in (B_\varepsilon(p) \cap A) \subset (B_\varepsilon(p) \cap (A \cup B))$.  A similar argument applies to $p \in B'$.
With this, the asserted equality becomes straightforward (noting that any set is a subset of its closure):
(1) $p \in \overline{A \cup B} \implies p \in \overline A \cup \overline B$.  Since $\overline S = S \cup S'$, either $p \in (A \cup B)$ or $p \in (A \cup B)'$.  In the first case, $p$ is immediately in $\overline A \cup \overline B$, and in the second case, our lemma shows this as well.
(2) $p \in \overline A \cup \overline B \implies p \in \overline{A \cup B}$.  Without loss of generality, assume $p \in \overline A = A \cup A'$.  $A$ is clearly a subset of $\overline{A \cup B}$, and our lemma shows $A' \subset (A \cup B)' \subset \overline{A \cup B}$ as well.
A: Lemma: $(A \cup B)' = A' \cup B'.$
Proof: Clearly, if $p$ is a limit point of $A$ or of $B$, then $p$ is a limit point of $A \cup B$.  We now prove the converse: Let $p \in (A \cup B)'$.  If for any $\varepsilon > 0$ there exists $q \neq p$ such that $q\in B_\varepsilon(p) \cap A$, then $p$ is a limit point of $A$.
Otherwise, there exists an $\varepsilon > 0$ where no such $q$ exists.  Then for any $\delta$ such that $0 < \delta < \varepsilon$, $B_\delta(p) \subset B_\varepsilon(p)$, and thus no $q \neq p$ exists in $B_\delta(p) \cap A$.  But for any such $\delta$, a $q \neq p$ must exist in $B_\delta(p) \cap (A \cup B)$, since $p$ is a limit point of $A \cup B$.  Consequently, $q \in B_\delta(p) \cap B$ for all such $\delta$, and $p$ is a limit point of $B$.
From this lemma, we have $\overline{A \cup B} = (A \cup B) \cup (A \cup B)' = A \cup B \cup A' \cup B' = \overline A \cup \overline B.$

Discussion: The proof has two components of note:

*

*Recognizing that the contribution is entirely in the lemma, and, once separated out, the entire proof follows immediately from it.


*Making use of the fact that balls of decreasing radius are subsets of each other.  I maintain that without this fact, the proofs above are not correct.
