[proofplan] The proof is an unpacking of what it means for an inhabited subset to be located. Locatedness supplies, for each point $x \in X$, a distance value $\operatorname{dist}(x,A)$ together with approximating points of $A$ within any prescribed positive rational tolerance. Applying this approximation clause with the given tolerance $\varepsilon$ gives the required point $a \in A$. The lower-bound clause in the definition records that the chosen point is approximating the actual located distance, not merely an arbitrary upper estimate. [/proofplan] [step:Fix the point and the rational tolerance] Let $x \in X$ be a point, and let $\varepsilon \in \mathbb{Q}$ satisfy $\varepsilon > 0$. Since $A$ is located, the distance from $x$ to $A$ is the real number $\operatorname{dist}(x,A)$ equipped with the located-distance approximation property: for every rational $\eta > 0$, there exists a point $a_\eta \in A$ such that \begin{align*} d(x,a_\eta) < \operatorname{dist}(x,A) + \eta. \end{align*} [/step] [step:Apply locatedness with tolerance $\varepsilon$] Apply the approximation property from locatedness with $\eta := \varepsilon$. Since $\varepsilon > 0$, this gives a point $a_\varepsilon \in A$ such that \begin{align*} d(x,a_\varepsilon) < \operatorname{dist}(x,A) + \varepsilon. \end{align*} Set $a := a_\varepsilon$. Then $a \in A$ and the required inequality holds. [guided] We now use the defining content of locatedness at the specific point $x$. The subset $A$ being located means more than the formal expression $\operatorname{dist}(x,A)$ existing as a symbol: it includes an approximation clause saying that any positive rational tolerance can be realized by an actual point of $A$. Here the prescribed tolerance is $\varepsilon \in \mathbb{Q}$, and the hypothesis gives $\varepsilon > 0$. Therefore the located-distance approximation clause applies with $\eta := \varepsilon$. It produces a point $a_\varepsilon \in A$ satisfying \begin{align*} d(x,a_\varepsilon) < \operatorname{dist}(x,A) + \varepsilon. \end{align*} Define $a := a_\varepsilon$. This point belongs to $A$ because the approximation clause produces witnesses inside $A$, and it satisfies exactly the inequality required in the theorem: \begin{align*} d(x,a) < \operatorname{dist}(x,A) + \varepsilon. \end{align*} No minimum of the set of distances $\{d(x,b) : b \in A\}$ is being assumed. The argument only uses the constructive approximation datum contained in locatedness. [/guided] [/step] [step:Record that the point is a genuine near-minimizer] The lower-bound part of the located-distance datum says that, for every $b \in A$, \begin{align*} \operatorname{dist}(x,A) \leq d(x,b). \end{align*} Thus the chosen point $a \in A$ lies within the prescribed rational error $\varepsilon$ above the located lower bound $\operatorname{dist}(x,A)$. This completes the proof. [/step]