Point in Finite Hausdorff Space is Isolated

Theorem
Let $T = \left({A, \vartheta}\right)$ be a Hausdorff 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 $X = \left\{{x_1, x_2, \ldots, x_n}\right\}$.

Let $x_k \in X$.

From the definition of Hausdorff space, $\forall x_i, x_j \in X: x_i \ne x_j: \exists U, V \in \vartheta: x_i \in U, x_j \in V: U \cap V = \varnothing$.

So there is a neighborhood of $x_k$ containing only $x_k$, so by definition, $x_k$ is an isolated point.

Hence the result.