[proofplan] We use the defining basis description of the metric topology: the sets $B_d(a,s)$ with $a \in X$ and $s > 0$ form a basis. If $U$ is open, then every point of $U$ lies in some basis element contained in $U$, giving the required ball. Conversely, if every point of $U$ contains such a ball, then $U$ is exactly the union of all those balls, hence is open because it is a union of basis elements. [/proofplan]

[step:Unpack openness in the metric topology to obtain a ball around each point] Let $\tau_d$ denote the metric topology induced by $d$ on $X$. By definition, the collection \begin{align*} \mathcal{B}_d := \{B_d(a,s) : a \in X \text{ and } s > 0\} \end{align*} is a basis for $\tau_d$. Assume $U \subset X$ is open in $\tau_d$, and let $x \in U$. Since $\mathcal{B}_d$ is a basis for $\tau_d$, there exist $a \in X$ and $s > 0$ such that \begin{align*} x \in B_d(a,s) \subset U. \end{align*} Because $x \in B_d(a,s)$, we have $d(a,x) < s$. Define \begin{align*} r := s - d(a,x). \end{align*} Then $r > 0$. If $y \in B_d(x,r)$, the triangle inequality gives \begin{align*} d(a,y) \le d(a,x) + d(x,y) < d(a,x) + r = s. \end{align*} Thus $y \in B_d(a,s)$, so $B_d(x,r) \subset B_d(a,s) \subset U$. Therefore, for this $x \in U$, there exists $r > 0$ such that $B_d(x,r) \subset U$. [/step]

[step:Write the set as a union of the balls contained in it]Assume that for every $x \in U$ there exists $r > 0$ such that $B_d(x,r) \subset U$. Define the indexing family \begin{align*} \mathcal{A} := \{B_d(x,r) : x \in U,\ r > 0,\ B_d(x,r) \subset U\}. \end{align*} We claim that \begin{align*} U = \bigcup_{A \in \mathcal{A}} A. \end{align*} First, if $z \in \bigcup_{A \in \mathcal{A}} A$, then $z \in A$ for some $A \in \mathcal{A}$. By the definition of $\mathcal{A}$, every such $A$ satisfies $A \subset U$, so $z \in U$. Conversely, let $z \in U$. By the assumed ball condition, there exists $\rho > 0$ such that $B_d(z,\rho) \subset U$. Hence $B_d(z,\rho) \in \mathcal{A}$, and since $d(z,z)=0<\rho$, we have $z \in B_d(z,\rho)$. Therefore $z \in \bigcup_{A \in \mathcal{A}} A$.[/step]

[guided]We want to prove that $U$ is open, and the metric topology is built from open balls. The natural way to connect the hypothesis to openness is to express $U$ as a union of open balls already known to lie inside $U$. Define \begin{align*} \mathcal{A} := \{B_d(x,r) : x \in U,\ r > 0,\ B_d(x,r) \subset U\}. \end{align*} This family contains exactly the open balls whose centers are points of $U$ and whose whole contents remain inside $U$. We prove the equality \begin{align*} U = \bigcup_{A \in \mathcal{A}} A. \end{align*} For the inclusion from right to left, suppose $z \in \bigcup_{A \in \mathcal{A}} A$. Then $z \in A$ for some $A \in \mathcal{A}$. By the defining property of $\mathcal{A}$, this particular set $A$ is contained in $U$. Hence $z \in U$. For the inclusion from left to right, suppose $z \in U$. The hypothesis says that there exists $\rho > 0$ such that \begin{align*} B_d(z,\rho) \subset U. \end{align*} Therefore $B_d(z,\rho)$ belongs to $\mathcal{A}$. Since $d(z,z)=0<\rho$, the center $z$ lies in its own open ball: \begin{align*} z \in B_d(z,\rho). \end{align*} Thus $z$ lies in one of the sets being unioned, so \begin{align*} z \in \bigcup_{A \in \mathcal{A}} A. \end{align*} The two inclusions prove the desired union representation.[/guided]

[step:Conclude openness from the union representation] Each element of $\mathcal{A}$ is an open ball, hence is a basis element for the metric topology $\tau_d$. Therefore each element of $\mathcal{A}$ is open in $\tau_d$. Since arbitrary unions of open sets are open in a topology, the equality \begin{align*} U = \bigcup_{A \in \mathcal{A}} A \end{align*} shows that $U$ is open in $\tau_d$. If $U=\varnothing$, the quantified ball condition is vacuous and $\varnothing$ is open by the axioms of a topology. This completes both implications. [/step]