A finite Hausdorff space is discrete

Let $X$ be a finite Hausdorff space. Let $x\in X$. For each $y\not =x\in X$ let $U_y$ and $V_y$ be disjoint open sets with $x\in U_x$ and $y \in V_y$. Set $V=\cup_{y\not = x} V_y$. Then $V$ is open... So $X\setminus V=\{x\}$ is closed.

Thus every point in $X$ is closed. Since $X$ is finite, every point is also open (complement of finite union of closed sets).

We could actually say that every finite $T_1$ space is discrete, since points being closed is equivalent to being $T_1$.

You're a bit sloppy in assuming that $\{x\}$ is open.

The thing you have to prove is that any subset of $X$ is open. This is quite straight forward as every subset of $X$ is $X$ minus a finite number of points, if it's not $X$ itself (which is open anyway) it's minus a finite positive number of points. That is you can write the subset as a finite intersection:

$$\bigcap X\setminus \{x_j\}$$

but the set $X\setminus \{x_j\}$ is open as you pointed out. And it's known that finite intersection of open sets is open. So any subset of $X$ is therefore open.

The same reasoning can be used to specially prove that $\{x\}$ is open, but we can prove the topology to be discrete directly here.

Yes, you're completely right (although, you might want to write out your argument for $\{x\}$ being open in a little more detail.) You've shown that every point is open, so it follows form the axioms for a topology that every set, as a union of points, is open. This is precisely the definition of the discrete topology.

Finiteness is used to conclude that every point is open. (It is certainly not true that every infinite Hausdorff space is discrete. Think of $\mathbb R$!) If you write out the argument more carefully, you'll see where finiteness is used.