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

Theorem
Let $$M = \left({A, d}\right)$$ be a metric space.

Let $$U \subseteq M$$ be a finite subset of $$M$$.

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

Then $$N_{D/2} \left({x_j}\right)$$ is a neighborhood of $$x_j$$ containing only $$x_j$$.

The result follows.