Open sets form a topology

Let $(X,d)$ be a metric space, and define

\tau_d=\{U\subseteq X:\text{for every }x\in U\text{ there is }r>0\text{ with }B_d(x,r)\subseteq U\},

where $B_d(x,r)$ is an open ball. Then:

$\varnothing,X\in\tau_d$ .
Any union of members of $\tau_d$ belongs to $\tau_d$ .
Any finite intersection of members of $\tau_d$ belongs to $\tau_d$ .

Thus $\tau_d$ is a topology on $X$ .

Proof

The empty set and $X$ lie in $\tau_d$ . An arbitrary union of members lies in $\tau_d$ , since each point belongs to one of them and has a ball inside it. For a finite intersection, take the minimum of the finitely many positive radii supplied at a point. The empty intersection is $X$ .

Resulting structure

This is the metric-induced topology. The proof starts with the metric ball condition, rather than assuming an existing topology and its open sets.