Proof of Euclidean Distance is a Metric on $\mathbb{R}^n$

Let $x=(x_1,\dots,x_n)$, $y=(y_1,\dots,y_n)$, and $z=(z_1,\dots,z_n)$ be points of $\mathbb{R}^n$. We verify the four conditions in Metric Space for the function $d_E$ from Euclidean Distance on $\mathbb{R}^n$.

First, each square $(x_i-y_i)^2$ is nonnegative, so the sum

is nonnegative. Hence its nonnegative square root is defined by Existence and Uniqueness of the Nonnegative Square Root, and therefore

Second, if $x=y$, then each difference $x_i-y_i$ is $0$, so $d_E(x,y)=0$. Conversely, if $d_E(x,y)=0$, then

Since each term in this sum is nonnegative, every term must be $0$. Thus $x_i=y_i$ for every $i\in\{1,\dots,n\}$, and hence $x=y$.

Third, for every $i$ one has

so

Finally, the triangle inequality for $d_E$ is the standard Euclidean inequality

Therefore

All four metric axioms hold, so $d_E$ is a metric on $\mathbb{R}^n$.