Point in Finite Metric Space is Isolated

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

Let $$X \subseteq A$$ such that $$X$$ is finite.

Then all points of $$X$$ are isolated.

Proof
As $$X$$ is finite, its elements can be placed in one-to-one correspondence with the elements of $$\N^*_n$$ for some $$n \in \N$$.

So let $$X = \left\{{x_1, x_2, \ldots, x_n}\right\}$$.

Now let $$\epsilon = \min \left\{{\forall i, j \in \N^*_n: i \ne j: d \left({x_i, x_j}\right)}\right\}$$.

That is, $$\epsilon$$ is the minimum distance between any two elements of $$X$$.

We have that $$\epsilon > 0$$ from the definition of metric.

Let $$N_{\epsilon} \left({x_k}\right)$$ be the $\epsilon$-neighborhood of any $$x_k \in X$$.

By the method of construction of $$\epsilon$$, it is clear that $$N_{\epsilon} \left({x_k}\right) = \left\{{x_k}\right\}$$.

Hence $$x_k$$ is an isolated point.

Alternative Proof
Alternatively, it can be noted that as a Metric Space is Hausdorff, we can apply the result All Points in Finite Hausdorff Space are Isolated‎.