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?

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 $x$.
