Proving the Metric Space Topology is a Topology Let $(X, d)$ be a metric space. Define $\tau = \left \{ u \subseteq X  | \exists S \subseteq B \ such\ that \ u = \bigcup_{s\in S}s  \right \}$, where $B$ is the set of all open balls in $X$. 
Show that $\tau$ is a topology on $X$.
I can show easily that $\phi \in \tau$ and that $X \in \tau$.
I've tried to prove that if $A\subseteq \tau$, then $\bigcup_{a\in A}a \in \tau$, but I'm not sure if it is correct and would appreciate verification. Here is my proof:
Let $A\subseteq \tau$ We want to find $S\subseteq B$ such that $\bigcup_{a \in A}a=\bigcup_{s \in S}s$.
From the definition of $\tau$, we know that for each $a \in A$ there exists $S_a \in B$ such that $a=\bigcup_{s\in S_a}s$.
Choose $S = \bigcup_{a \in A}S_a$ This is a subset of $B$ as each $S_a$ is a subset of $B$
Let $x\in \bigcup_{a \in A}a$, then $x \in a_i$ for at least some $a_i \in A$.
But, $\bigcup_{s \in S}s=(\bigcup_{s \in S_{a_i}}s)\cup(\bigcup_{s \in comp(S_{a_i})}s)=(a_i)\cup(\bigcup_{s \in comp(S_{a_i})}s)$
so $a_i \subseteq \bigcup_{s \in S}s$, thus $\bigcup_{a \in A}a \subseteq\bigcup_{s \in S}$
Now let $x\in \bigcup_{s \in S}s$, then $x \in b$ for at least one $b \in S$
Recall, $S = \bigcup_{a \in A}S_a$, so $b\in S_{a_i}$, for some $a_i\in A$
But, $a_i=\bigcup_{s\in S_{a_i}}s$, so $b \subseteq a$ and thus $b \subseteq \bigcup_{a \in A}a$ so $x \in \bigcup_{a \in A}a$. Thus,  $\bigcup_{a \in A}a \supseteq\bigcup_{s \in S}$
So we have shown $\bigcup_{a \in A}a =\bigcup_{s \in S}$, as required.
Finally I'm trying to prove that if $u_1,u_2 \in \tau$, then $(u_1 \cap u_2)\in \tau$. I can't seem to figure it out, but here is my progress so far:
As $u_1,u_2 \in \tau$, there exists $S_1,S_2\subseteq B$ such that $u_1 = \bigcup_{s\in S_1}s$, and $u_2 = \bigcup_{s\in S_2}s$.
We need to find $S\subseteq B$ such that $(\bigcup_{s\in S_1}s) \cap (\bigcup_{s\in S_2}s)=\bigcup_{s\in S}s$
Intuitively I want $S$ to consist of an open ball centred on each point in $u_1 \cap u_2$ of small enough radius that the ball is contained entirely in $u_1 \cap u_2$, but I have been unable to translate this idea into maths.
Any help with the verification of part 3 or any ideas/help for the proof of part 3 would be greatly appreciated!
 A: I don't like your choice of notation. It's better to use upper cases for sets. Having said that let's observe that $$\tau=\{U\subseteq X|\ \forall x\in U:\exists B\in\mathcal{B}:x\in B\subseteq U\}$$, where $\mathcal{B}$ is the family of all open balls in $(X,d)$. Now all you need to prove is that $\mathcal{B}$ is a base for $\tau$, i.e.


*

*For each $x\in X$ , there exists $B\in\mathcal{B}$ such that $x\in B$;

*For each $B_1,B_2\in\mathcal{B},$ if $x\in B_1\cap B_2,$ then there exists $B_3\in \mathcal{B}$ such that $x\in B_3\subseteq B_1\cap B_2$.


$U_1,U_2\in \tau\implies U_1\cap U_2\in \tau$ is immediate now.
A: Your notation could be better.
$O$ is open iff $$O = \bigcup \{B(x_i, r_i): i \in I(O)\}$$
for some set of centres $x_i$ and radii $r_i$ for some index set $i \in I(O)$.
Now, if we write $U$ and $V$ as such unions for index sets $I(U), I(V)$.
For every $y \in U \cap V$ we have $i_1 \in I(U)$ and $x_{i_1}, r_{i_1}$ and $i_2 \in I(V)$ and $x_{i_2}, r_{i_2}$ such that 
$$y \in B(x_{i_1}, r_{i_1}) \cap B(x_{i_2}, r_{i_2})$$
Now, the triangle inequality says that if we take $r(y)=\min(r_{i_1}-d(y, x_{i_1}, r_{i_2}-d(y, x_{i_2})$, we have 
$$B(y, r(y) \subseteq B(x_{i_1}, r_{i_1}) \cap B(x_{i_2}, r_{i_2})$$
so that 
$$U \cap V = \bigcup \{B(y, r(y): y \in U \cap V\}$$
which is open as a union of balls. Proving we have a (local) base with the balls is easier though. But  I thought I'd give a proof along your lines.
As to unions, we write $O_j, j \in J$ as a union as in the beginning with index set $I(O_j)$. Then with the index set $I = \bigcup_j I(O_j)$ we can write 
$$ \bigcup O_j = \bigcup_{i,j} \{B(x_i, r_i) :i \in I(O_j) \}$$
