[proofplan] We verify directly that the positive-radius open balls satisfy the basis axioms. First, every point of $X$ lies in an open ball. Second, whenever a point lies in the intersection of two open balls, the triangle inequality produces a smaller open ball around that point contained in the intersection. Finally, we identify the topology generated by this basis with the metric topology by using the defining local $\varepsilon$-ball condition for metric openness. [/proofplan]

[step:Show the open balls cover the metric space] Let $\tau_d$ denote the metric topology on $X$. For every $x \in X$, the ball $B_d(x,1)$ belongs to $\mathcal{B}_d$ because $1 > 0$, and $x \in B_d(x,1)$ because $d(x,x)=0<1$. Hence \begin{align*} X = \bigcup_{B \in \mathcal{B}_d} B. \end{align*} Thus $\mathcal{B}_d$ covers $X$. [/step]

[step:Refine intersections of open balls by a smaller open ball]Let $B_d(x_1,r_1), B_d(x_2,r_2) \in \mathcal{B}_d$, and let $y \in B_d(x_1,r_1) \cap B_d(x_2,r_2)$. Define the positive number \begin{align*} \varepsilon := \frac{1}{2}\min\{r_1 - d(x_1,y), r_2 - d(x_2,y)\}. \end{align*} The two quantities inside the minimum are positive because $y \in B_d(x_1,r_1)$ and $y \in B_d(x_2,r_2)$, so $\varepsilon > 0$. We claim that $B_d(y,\varepsilon) \subset B_d(x_1,r_1) \cap B_d(x_2,r_2)$. Let $z \in B_d(y,\varepsilon)$. By the triangle inequality, \begin{align*} d(x_1,z) \le d(x_1,y) + d(y,z) < d(x_1,y) + \varepsilon \le d(x_1,y) + r_1 - d(x_1,y) = r_1. \end{align*} Thus $z \in B_d(x_1,r_1)$. The same argument with $x_2$ and $r_2$ gives \begin{align*} d(x_2,z) \le d(x_2,y) + d(y,z) < d(x_2,y) + \varepsilon \le d(x_2,y) + r_2 - d(x_2,y) = r_2. \end{align*} Thus $z \in B_d(x_2,r_2)$, proving the claimed containment.[/step]

[guided]We must prove the local refinement condition for a basis: given two basis candidates and a point in their intersection, we need a third basis candidate around that point and contained in the intersection. Let the two balls be $B_d(x_1,r_1)$ and $B_d(x_2,r_2)$, where $x_1,x_2 \in X$ and $r_1,r_2 > 0$. Let $y \in B_d(x_1,r_1) \cap B_d(x_2,r_2)$. Since $y$ lies in the first ball, $d(x_1,y) < r_1$, so $r_1 - d(x_1,y) > 0$. Since $y$ lies in the second ball, $d(x_2,y) < r_2$, so $r_2 - d(x_2,y) > 0$. Define \begin{align*} \varepsilon := \frac{1}{2}\min\{r_1 - d(x_1,y), r_2 - d(x_2,y)\}. \end{align*} This is positive, so $B_d(y,\varepsilon)$ is an element of $\mathcal{B}_d$. Now take any $z \in B_d(y,\varepsilon)$. The choice of $\varepsilon$ ensures that $d(y,z) < \varepsilon \le r_1 - d(x_1,y)$. Applying the triangle inequality in the [metric space](/page/Metric%20Space) $(X,d)$ gives \begin{align*} d(x_1,z) \le d(x_1,y) + d(y,z) < d(x_1,y) + r_1 - d(x_1,y) = r_1. \end{align*} Therefore $z \in B_d(x_1,r_1)$. The same calculation uses the other available margin, namely $\varepsilon \le r_2 - d(x_2,y)$. Again by the triangle inequality, \begin{align*} d(x_2,z) \le d(x_2,y) + d(y,z) < d(x_2,y) + r_2 - d(x_2,y) = r_2. \end{align*} Therefore $z \in B_d(x_2,r_2)$. Since every $z \in B_d(y,\varepsilon)$ lies in both original balls, we have \begin{align*} B_d(y,\varepsilon) \subset B_d(x_1,r_1) \cap B_d(x_2,r_2). \end{align*} This is exactly the intersection-refinement axiom for a basis.[/guided]

[step:Conclude that the generated topology is the metric topology] The previous two steps show that $\mathcal{B}_d$ is a basis for some topology on $X$, namely the topology whose open sets are arbitrary unions of elements of $\mathcal{B}_d$. It remains to identify this topology with $\tau_d$. First, every ball $B_d(x,r)$ is open in $\tau_d$: if $y \in B_d(x,r)$, then with \begin{align*} \delta := r - d(x,y) \end{align*} we have $\delta > 0$, and the triangle inequality gives $B_d(y,\delta) \subset B_d(x,r)$. Hence every union of balls belongs to $\tau_d$. Conversely, let $U \in \tau_d$. By the defining local property of metric openness, for each $x \in U$ there exists $\varepsilon_x > 0$ such that $B_d(x,\varepsilon_x) \subset U$. Therefore \begin{align*} U = \bigcup_{x \in U} B_d(x,\varepsilon_x), \end{align*} where each $B_d(x,\varepsilon_x)$ belongs to $\mathcal{B}_d$. Thus every metric-[open set](/page/Open%20Set) is a union of elements of $\mathcal{B}_d$. The topology generated by $\mathcal{B}_d$ is therefore exactly $\tau_d$, so $\mathcal{B}_d$ is a basis for the metric topology on $X$. [/step]