Prove equivalence between $X$ Hausdorff and $X$ finite with discrete topology We have a Noetherian topological space $X$. Show that the following are equivalent:


*

*$X$ is a Hausdorff space

*$X$ 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 ($1\implies 2$). 
 A: 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.
A: $X$ 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.
