The union of two open sets is open (In metric Spaces) Let $X$ a set not empty and $(X,d)$ a metric space. Prove he union of two open sets is open.
My proof:


Let $A_1,A_2$ open sets, we need to prove $A_1\cup A_2$ is open.
  As $A_1,A_2$ are open set, then for all $a_1,a_2\in A_1,A_2$ respectively we have $r_1,r_2>0$ such that $B(a_1,r_1)\subset A_1$ and $B(a_2,r_2)\subset A_2$
  Let $r=\frac{1}{2}min\{r_1,r_2\}$ and $x\in A_1\cup A_2$, then $x \in A_1$ or $x \in A_2$
  This implies: $B(x,r)\subset A_1\cup A_2$
  In conclusion, $A_1\cup A_2$  is open set.
  

Note: My definition of Open is $\forall x\in A_1$ exists $r>0$ such that $B(x,r)\subset A$
What is your opinion about my proof? Do you think is a good proof? Is convincing?
 A: The general idea is correct. The problem is that the radius $r$ depends on the given point. It seems to me as you use that the radius is independent of the given point which is not true. Hence, I would write it down differently.
My solution:
Let $x \in A_1 \cup A_2$ be arbitrary. Without loss of generality we can assume $x \in A_1$. As $A_1$ is open, there exists $r>0$ such that $B(x,r)\subset A_1\subset A_1 \cup A_2$. Hence, $A_1 \cup A_2$ is open.
A: Nope. It isn't a good proof. Note that $r_1, r_2$ depends on the elements $a_1,a_2$. There is a simpler proof: Take $x \in A_1 \cup A_2$. Then $x \in A_1$ or $x \in A_2$. If $x \in A_1$, as $A_1$ is open, there exists an $r>0$ such that $B(x,r) \subset A_1 \subset A_1 \cup A_2$ and $B(x,r)$ is an open set. 
Therefore $A_1 \cup A_2$ is an open set.
A: Considering $a_1,a_2$ is confusing. In fact, the union of any collection of open sets is open. Let $A=\bigcup_t A_t$, where all $A_t$ are open. Let $x\in A$. So, there is at least one $t$ for which $x\in A_t$. Therefore for some $r>0$ we have $B(x,r)\subset A_t\subset\bigcup_t A_t=A$.
This proof works in any topological space, not necessarily metric. It is enough to replace a ball $B(x,r)$ with an open neighbourhood of $x$.
