Proof of Euclidean Openness Agrees with Metric Openness on $\mathbb{R}^n$

Let $U\subseteq \mathbb{R}^n$.

Assume first that $U$ is open in the Euclidean sense. Let $x=(x_1,\dots,x_n)\in U$. By Open Subset of Euclidean Space, there exists a real number $r>0$ such that every point $y=(y_1,\dots,y_n)\in\mathbb{R}^n$ satisfying

$$\sum_{i=1}^n (y_i-x_i)^2<r^2$$

belongs to $U$. By the definition of the Euclidean distance, the inequality above is equivalent to

$$d_E(x,y)<r.$$

Thus

$$B_{d_E}(x,r)\subseteq U,$$

where $B_{d_E}(x,r)$ is the open ball in the metric space $(\mathbb{R}^n,d_E)$. Since $x\in U$ was arbitrary, $U$ is open in the metric space $(\mathbb{R}^n,d_E)$.

Conversely, assume that $U$ is open in the metric space $(\mathbb{R}^n,d_E)$. Let $x=(x_1,\dots,x_n)\in U$. Then there exists $r>0$ such that

$$B_{d_E}(x,r)\subseteq U.$$

If $y=(y_1,\dots,y_n)\in\mathbb{R}^n$ satisfies

$$\sum_{i=1}^n (y_i-x_i)^2<r^2,$$

then by the definition of $d_E$ one has $d_E(x,y)<r$, so $y\in B_{d_E}(x,r)\subseteq U$. Therefore $U$ is open in the Euclidean sense.

Hence $U$ is open in the Euclidean sense if and only if it is open in the metric space $(\mathbb{R}^n,d_E)$.