Finite Subspace of Dense-in-itself Metric Space is not Open

Theorem
Let $M = \left({A, d}\right)$ be a metric space that is dense-in-itself.

Let $U$ be a finite subset of $A$.

Then $U$ is not an open set of $M$.

Proof
Let $U = \left\{{x_0, x_1, \ldots, x_n}\right\}$.

Let $x_j \in U$.

Let $\displaystyle D = \min_{i \ne j} d \left({x_i, x_j}\right)$.

Then $B_{D/2} \left({x_j}\right)$ is a open ball of $x_j$ containing only $x_j$.

The result follows.