What is wrong with this proof of the union of two compact sets? I need to show that the union of two compact sets is compact. Here is what I tried:
Let $A$ and $B$ be two compact sets. Then for every open cover $\{A_n\}_{n\in I_1}$ and open cover $\{B_m\}_{m\in I_2}$, there exists finite subcovers $\bigcup\limits_{n\in I_1}^p A_n$ such that $A\subseteq\bigcup\limits_{n\in I_1}^p A_n$ and $\bigcup\limits_{b\in I_2}^k B_m$ such that $B\subseteq\bigcup\limits_{b\in I_2}^k B_m$. Then, $A\cup B\subseteq (\bigcup\limits_{n\in I_1}^p A_n)\cup(\bigcup\limits_{b\in I_2}^k B_m)\subseteq(\bigcup\limits_{n\in I_1} A_n) \cup (\bigcup\limits_{b\in I_2} B_m)$ (the union of open covers constructed as an open cover for $A\cup B$). Thus, for ever open cover of $A\cup B,$ there exists a finite subcover, so $A\cup B$ is compact.
I thought that since the open cover I constructed was arbitrary, and so was the finite subcover, then every open cover of $A\cup B$ had a finite subcover. What is wrong with this?
 A: $$A\cup B \subseteq \left( \bigcup_{n \in I_1} A_n \right)\cup \left( \bigcup_{m \in I_2} B_m \right)$$
may not be a general open cover for $A\cup B$. So, it is better to start with:
Let $\{ C_n \}$ be an open cover for $A \cup B$. Then since both $A$ and $B$ are compact sets, we have $\dots$
A: The problem with your logic is as follows:
You are right, in that a finite subcover of $A$ and a finite subcover of $B$, together gives a finite subcover of $A \cup B$.
However, to show $A \cup B$ is compact, you are supposed to start with any arbitrary open cover of $A \cup B$. This was not done: instead, you started with open covers for $A$ and $B$ separately, and used their compactness to produce a finite subcover.

The missing step : How do you know that a subcover of $A \cup B$, is also a subcover of $A$ and a subcover of $B$? Or, how would you generate subcovers of $A$ and $B$, given one of $A \cup B$? In short : you did not start with an arbitrary subcover. If you do so, it would have to be as explicit as in the answer I am going to give below for your betterment.

Hence, the answer should be as follows:
Start from definition. Let $\{ U_\alpha\}$ be an open cover of $A \cup B$.
Since $A \cup B \subset \bigcup U_\alpha$, it follows that $A \subset \bigcup U_\alpha$, and similarly, $B \subset \bigcup U_\alpha$. That is, every cover of $A \cup B$ gives a cover of $A$ and a cover of $B$. This is the key step missing from your explanation.
Now, we have finite subcovers for each one, and you can take the union of these subcovers (which remains a subcover of the original cover) to get the result.
So, in short, when we say arbitrary, we have to be careful. It is best that you start out with what is given: In compactness, we are given an open cover of the set, you never gave yourself that, instead you started with open covers for $A$ and $B$ separately, which was not given to you by the definition of $A \cup B$ being compact.
It's all about being careful. And it's absolutely fine to mistakes. It's good you have this clarified now.
As an exercise, try and extend this to a finite union of compact sets. Do you think you can extend this to an infinite union of compact sets?
A: You assumed, without proving it, that every open cover of $A\cup B$ is a union of an open cover of $A$ and an open cover of $B$.
Suppose for example, that $A=[0,1]$ and $B$ is some other compact set, and $\mathcal B = \{ (0.4,1] \cup G : G\text{ is an open subset of } B. \}.$ This is an open cover of $B$. Let $C=\{ [0,0.6)\cup G : G \text{ is an open subset of } B. \}.$ That is another open cover of $B$. Neither of them is an open cover of $A,$ but $\mathcal B \cup \mathcal C$ is an open cover of $A\cup B.$ Each has a finite subset that covers $B.$ Neither has a finite subset that covers $A$.
So your proof is incomplete.
You could just say that any open cover of $A\cup B$ is an open cover of $A$. So it has a finite subcover of $A$. Similarly find another finite subset of it that covers $B$. Then take the union of those two finite subsets. That actually makes the proof simpler. If you do it that way then there is no need to prove the proposition whose proof you omitted.
PS: The needless assumption whose proof you omitted can be proved trivially: If $\mathcal A$ is an open cover of $A\cup B$ then $\mathcal A \cup \mathcal A$ is a union of an open cover of $A$ and an open cover of $B$. However, this is a needless complication in the argument.
