Prove that if $A$ and $B$ are subsets of a topological space $(X, \mathcal{T})$, then $\overline{A \cup B} = \overline{A} \cup \overline{B}$ 
Prove that  if $A$ and $B$ are subsets of a topological space $(X, \mathcal{T})$, then $\overline{A \cup B} = \overline{A} \cup \overline{B}$


My Proof
We'll use the notation, $U_x$ to be an open set of $X$ containing $x$, (i.e $U_x := x \in U$ and $U \in \mathcal{T})$. 
Now $x \in \overline{A \cup B}$ if every $U_x \cap (A \cup B) \neq \emptyset \iff (U_x \cap A) \cup (U_x \cap B) \neq \emptyset$. (The logical equivalence here is as a result of elementary set operations).
And we have $x \in \overline{A}$ if $U_x \cap A \neq \emptyset$ for every $U_x \in \mathcal{T}$, and similarly  $x \in \overline{B}$ if $U_x \cap B \neq \emptyset$ for every $U_x \in \mathcal{T}$, so that $x \in \overline{A} \cup \overline{B}$ if $(U_x \cap A \neq \emptyset) \lor (U_x \cap B \neq \emptyset).$
Logically we have $(U_x \cap A \neq \emptyset) \lor (U_x \cap B \neq \emptyset) \iff (U_x \cap A) \cup (U_x \cap B) \neq \emptyset$, and hence it follows by the biconditional  above that $x \in \overline{A \cup B} \implies  x \in \overline{A} \cup \overline{B}$ and that $x \in \overline{A} \cup \overline{B} \implies x \in \overline{A \cup B}$ and thus we have $\overline{A \cup B} = \overline{A} \cup \overline{B}$. $\square$

Is my proof correct? If so how rigorous is it? I'm looking to improve my proof-writing skills, and the rigor in my proofs, so any comments and criticism, however critical they may be is greatly appreciated.
 A: So far you have shown the inclusion $\overline{A\cup B}\subset\overline{A}\cup\overline{B}$. This proof is correct. Try to prove the converse inclusion now.
A: As  szw1710 has observed you need to show the reverse inclusion. In general, most proofs for sets that R=S involving showing that R is a subset of S and vice-versa. Let's look at the proof of the containment you have presented. First as a matter of style you should treat the reader (especially if he is your professor) as royalty. Your gracious highness let me introduce to you x who is a trusted member of  A∪B closure. Well that's probably overkill, but you do need to introduce x and state where he comes from. You are trying to show that x belongs to A closure or x belongs to B closure. Here there is a great opportunity to use a result from logic: P or Q is logically equivalent to if not P, then Q. So suppose x is  not a member of  A closure. To see that x belongs to B closure, let U be an open set containing x. You know that you are going to need to use that x is not in A closure and here is your chance. Since x is not in A closure, there is an open set V containing x such that...
The last line of the proof of this containment ought be something like, "Thus U∩B≠∅ and so ... 
In Answers I am supposed to provide an answer, and I obviously have not. So I am in danger of losing some of my reputations. But you can answer your own question along the lines I have suggested an rescue me. BTW I do not think the argument you have given is correct, because there is a difference between saying every open set about x hits A or B and saying x is in A closure or x is in B closure.     
A: You're making a common error: you seem to assume the following are logically equivalent (Interchanging $\forall$ and $\lor$):


*

*$\forall U \in \mathcal{U}_x: U \cap (A \cap B) = (U \cap A) \cup (U \cap B) \neq \emptyset$ so $\forall U \in \mathcal{U}_x: (U \cap A \neq \emptyset) \lor (U \cap B \neq \emptyset)$.

*$(\forall U \in \mathcal{U_x}: U \cap A \neq \emptyset )\lor (\forall U \in \mathcal{U_x}: U \cap B \neq \emptyset)$.
where $\mathcal{U}_x$ is the set of open sets that contain $x$.


*is what it means for $x$ to be in $\overline{A} \cup \overline{B}$, while 1 means that $x \in \overline{A \cup B}$.


Logically we can say that 2 implies 1, which gives us one inclusion.
To see that they actually are equivalent we need a basic fact about open $\mathcal{U_x}$: it's closed under intersections of two elements.
So suppose 2 does not hold: it means that there is an $U_1 \in \mathcal{U}_x$ such that $U_1 \cap A = \emptyset$ and also a $U_2 \in \mathcal{U}_x$ such that $U_2 \cap B = \emptyset$, but then $U_1 \cap U_2 \in \mathcal{U}_x$ and $(U_1 \cap U_2) \cap (A \cup B) = \emptyset$, so 1. fails too. So not 2 implies not 1 (so 1 implies 2). 
