Proof that open balls form a basis of the metric topology

Proof

We check the second form of the definition of a basis for both families.

1. Let $\mathcal{B}$ be the set of all open balls $B(x, r)$ with $x \in X$ and $r > 0$. Let $\mathcal{B}'$ be the set of balls $B(x, 1/n)$ with $x \in X$ and integer $n \ge 1$. Then $\mathcal{B}' \subseteq \mathcal{B}$.
2. By open balls are open, each member of $\mathcal{B}$ is in the metric topology $\tau_d$. So $\mathcal{B}' \subseteq \mathcal{B} \subseteq \tau_d$.
3. Let $U \in \tau_d$ and $x \in U$. By the definition of $\tau_d$, there is a real $\varepsilon > 0$ with $B(x, \varepsilon) \subseteq U$.
4. The ball $B(x, \varepsilon)$ is in $\mathcal{B}$ and contains $x$, since $d(x, x) = 0 < \varepsilon$. So $\mathcal{B}$ is a basis of $\tau_d$.
5. Since the reals are Archimedean, there is an integer $n \ge 1$ with $n \varepsilon > 1$, that is, $1/n < \varepsilon$.
6. If $d(x, y) < 1/n$, then $d(x, y) < \varepsilon$. So $x \in B(x, 1/n) \subseteq B(x, \varepsilon) \subseteq U$.
7. The ball $B(x, 1/n)$ is in $\mathcal{B}'$. So $\mathcal{B}'$ is also a basis of $\tau_d$.