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 $U = \left\{{x_0, x_1, \ldots, x_n}\right\}$, and assume for contradiction that $U$ is open.

Let $x_j \in U$.

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

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

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

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

However, since $D/2 < \displaystyle \min_{i \ne j} d \left({x_i, x_j}\right)$, it is clear that this open set is simply the singleton $\left\{{x_j}\right\}$.

But then $x_j$ is an isolated point of $A$, contradicting the fact that $M$ is dense-in-itself.

So in fact, $U$ cannot be open, and the result follows.