Yes. Let X be finite and Hausdorff. It is enough to show that every point x in X is open. For every point y different from x, there is an open neighborhood U_y of x not meeting y. The intersection of the U_y's is open and equals {x}.

Yes. Let $X$ be finite and Hausdorff. It is enough to show that every point $x$ in $X$ is open. For every point $y$ different from $x$, there is an open neighborhood $U_{y}$ of $x$ not meeting $y$. The intersection of the $U_{y}$'s is open and equals $\left\{ x\right\}$.