Prove equivalence between $X$ Hausdorff and $X$ finite with discrete topology

Question

We have a Noetherian topological space $X$. Show that the following are equivalent:

1. $X$ is a Hausdorff space
2. $X$ is finite and has discrete topology

So far I've only got this:

If $X$ has discrete topology, then any two points $x_1$ and $x_2$ are elements of the disjoint open sets $\{x_1\}$ and $\{x_2\}$ respectively. This implies $X$ is Hausdorff. (So 2$\implies1$)

I don't know how to prove ($1\implies 2$).

Answer 1

$X$ is a Noetherian topological space, so there exist a nonnegative integer $n$ and closed, irreducible subsets $Z_1,\ldots, Z_n$ such that $X=Z_1\cup\ldots\cup Z_n$, with $Z_i\not\subseteq Z_j$ for $i\neq j$.

Now suppose we have two distinct points $x,y\in Z_i$. The Hausdorff property of $X$ implies that we must now also be able to find disjoint open sets $U_1\ni x$ and $U_2\ni y$. In the subspace topology of $Z_i$ we now see that the two open sets $U_1\cap Z_i$ and $U_2\cap Z_i$ are disjoint. This contradicts the fact that $Z_i$ is irreducible. Hence we cannot find two distinct points in any $Z_i$, which in turn implies that $X$ is a finite union of singletons. Hence $X$ is a finite Hausdorff space which necessarily means it also has the discrete topology.

Answer 2

Assume first that $X$ is irreducible. Then every nonempty open $U$ subset of $X$ is dense. But if $X$ is Hausdorff the existence of a point in the complement of $U$ is impossible. Therefore every nonempty open subset of $X$ is $X$ itself. Again since $X$ is Hausdorff this implies that $X$ consists only of one point.

In general, since $X$ is noetherian you can write it as a finite union of its irreducible components $X_1,\ldots,X_n$. Since $X$ is Hausdorff the $X_i$'s are also Hausdorff and thus by the above consist of one point. Therefore $X$ is finite with $n$ points and every subset consisting of one point is closes. This means that $X$ has the discrete topology.