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

Aiming for a contradiction, suppose that $U$ is open.

Let $x_j \in U$.

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

Consider the open ball $B_{D/2} \left({x_j}\right)$ of $x_j$.

From Open Ball is Open Set, $B_{D/2} \left({x_j}\right)$ is an open set.

Then from Finite Intersection of Open Sets of Metric Space is Open, we can see that $U \cap B_{D/2}$ is also an open set.

However, we have that:
 * $D/2 < \displaystyle \min_{i \mathop \ne j} d \left({x_i, x_j}\right)$

Thus this open set is simply the singleton $\left\{{x_j}\right\}$.

But then $x_j$ is an isolated point of $A$.

This contradicts the fact that $M$ is dense-in-itself.

So in fact, $U$ cannot be open.

The result follows.