Proof of usual topology on $\mathbb{C}$ I want to show that

Let $\mathcal{T}$ be the set of all open sets in $\mathbb{C}$. Then:
  
  
*
  
*$\emptyset \in \mathcal{T}$ und $\mathbb{C} \in \mathcal{T}$
  
*For arbitrary many $U_i \in \mathcal{T}$ is: $(\bigcup_{i \in
   \mathcal{I}} U_i) \in \mathcal {T}$
  
*For finitely many $U_1,\ldots,U_n \in \mathcal{T}$ is: $(\bigcap_{i
   = 1}^n U_i) \in \mathcal {T}$


And I want to use this definition of openness in $\mathbb{C}$:

Let $U \in \mathbb{C}$.
$U$ is open $\iff \forall u \in U \,\exists r \gt 0: B_r(u) \subseteq U$
$U$ is closed $\iff U^C := \mathbb{C} \setminus U$ is open

And the open ball is defined as $B_r(u) := \{z \in \mathbb{C} \,|\, |u-z| \lt r\}$.

I get how the first point is vacuously true. But I struggle how to show the other two points. How do I make the distinction between finitely and (possibly) infinitely many open sets in the proof?
 A: For the second point:
$\forall x\in \bigcup_{i\in\mathcal I}U_i$, $x\in U_i$ for some $i\in\mathcal{I}$, so there exists an open ball $B_r(x)\subset U_i$ and thus $B_r(x)\subset \bigcup_{i\in\mathcal I}U_i$. Hence $\bigcup_{i\in\mathcal I}U_i$ is open.
For the third point:
$\forall x\in \bigcap_{i=1}^nU_i$, there are open balls $B_{r_i}(x)\subset U_i$. Let $r=\min_{1\leq i\leq n}r_i$, then $B_r(x)\subset \bigcap_{i=1}^nU_i$. Hence $\bigcap_{i=1}^nU_i$ is open.
A: Let $T$ be a topology on a set $X.$ A base (basis) for $T$ is some  (any) $B\subset T$ such that each non-empty $U\in T$ is the union of a family (finite or infinite) of (one or more) members of $B.$ Of course $T$ is itself an example of a base for $T.$  So  $B$ is a base for $T$ iff $T=\{\cup F: F\subset B\}.$
If $B$ is any family of subsets of $X$ then $B$ is a base for a topology on $X$ iff 
(Bi). $\cup B=X.$ (Each point of $X$ belongs to at least one member of $B.$),and 
(Bii). If $U,U'\in B$ and $p\in U\cap U'$ then there exists $U''\in B$ with $p\in 
 U''\subset (U\cap U').$ 
(Note that (Bii) is automatically satisfied if $U\cap U'\in B$ whenever $U,U'\in B,$ as we can let $U''=U\cap U.$ For example, with the usual topology on $\Bbb R,$ the base of bounded open intervals with rational endpoints has this property.)
A metric on a set $X$ is a function $d:X\times X\to [0,\infty)$ such that 
(Di). $d(x,y)=d(y,x),$ and 
(Dii). $d(x,y)=0$ iff $x=y,$ and
(Diii). $d(x,y)+d(y,z)\geq d(x,z).$ (Triangle Inequality).
If $d$ is a metric on $X$ then the set of open $d$-balls is a base for a topology on $X.$  Condition (Bi) is obviously satisfied. To satisfy (Bii) we use  (Diii).
The function $d(x,y)=|x-y|$ is a metric on $\Bbb C.$
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