Solution: Let $A = \{ x_1, x_2, \dots, x_n \} $ be a finite subset of $X$. We will prove that $A$ is closed. Since finite union of closed sets is closed, so it is sufficient to show that each singletons are closed. So let $A = \{ x_1 \} $. Take a point $x\in X$ different from $x_1$. If such a point does not exist, then $X$ is a one point set and hence closed and we are done. Since $X$ is Hausdorff, we can find two disjoint neighborhoods $U_{x_1}$ and $U_{x}$ of $x_1$ and $x$ respectively. Since $\{ x_0 \} $ does not intersects with $U_{x}$, therefore, $x$ can not belong to $\bar{A} $. Since $x$ is an arbitrary point, no points of $X$ other than $x_0$ can belong to $\bar{A} $ and therefore $\bar{A} =A$ Thus, $A$ is closed.