Proof of The Borel $\sigma$-Algebras of Euclidean Space and of the Euclidean Metric Coincide

By Euclidean Openness Agrees with Metric Openness on $\mathbb{R}^n$ a subset of $\mathbb{R}^n$ is open in the Euclidean sense if and only if it is an open subset of the metric space $(\mathbb{R}^n,d)$. Write $\mathcal{U}$ for this single family of subsets of $\mathbb{R}^n$: it is at once the family of all Euclidean open subsets of $\mathbb{R}^n$ and the family of all open subsets of $(\mathbb{R}^n,d)$.

By Borel Sigma-Algebra on Euclidean Space the $\sigma$-algebra $\mathcal{B}(\mathbb{R}^n)$ is the $\sigma$-algebra generated by the first of these families, and by Borel Sigma-Algebra of a Metric Space the $\sigma$-algebra $\mathcal{B}(\mathbb{R}^n,d)$ is the $\sigma$-algebra generated by the second; so both are the $\sigma$-algebra generated by $\mathcal{U}$. The $\sigma$-algebra generated by a family of subsets of $\mathbb{R}^n$ consists of those subsets of $\mathbb{R}^n$ that belong to every $\sigma$-algebra on $\mathbb{R}^n$ containing the family, so it is determined by the family alone. Hence $\mathcal{B}(\mathbb{R}^n,d)=\mathcal{B}(\mathbb{R}^n)$.