Hausdorff Spaces Subsets open? 'Given a Hausdorff space $X$ with finitely many elements, show all subsets of $X$ are open in $X$.'
I let $U$ be an arbitrary subset of $X$
Since every subset of a Hausdorff space is a Hausdorff space, then $U$ is also Hausdorff. 
This means that any two points in $U$ $x \ne y$ have disjoint neighbourhoods.
Does this imply every point of $U$ has a neighbourhood lying in $U$ hence prove open? This seems trivial. Where does finitely many elements come into it?
 A: Let $X=\{x_1,...,x_n\}$. Given $x_i$, for $j\ne i$ let $U_{i,j}$ be an open neighborhood of $x_i$ that doesn't contain $x_j$. Then $U_i = \bigcap_{j\text{:  } j\ne i}U_{i,j}$ is a finite intersection open neighborhood of $x_i$ that contains no other $x_j$, so it's open. But $U_i = \{x_i\}$.
So all singletons $\{x_i\}$ are open. Every subset of $X$ is a union of singletons, so every subset is open.
A: In a finite Hausdorff $X=\{x_1, \ldots, x_n\}$ space you can show that every singleton $\{x_i\}$ is open. Hence every subset is open since it is union of open sets. 
To see that $\{x_i\}$ is open, we know that for every $j\neq i$, we have disjoint neighborhoods $U_j, V_j$ of $x_i, x_j$ respectively.
Take $$U = \bigcap_{j\neq i} U_j$$
Then $U = \{x_i\}$, which is open since it is finite intersection of open sets.
A: We only need that singletons are closed (because then all finite sets are closed, and so all sets are open)...
A: Just for interest, here is a simpler solution to the question I asked. 
We want to prove any subset $U$ in a Hausdorff space with finitely many elements is open.
If $X$ is Hausdorff, then by definition every finite subset of $X$ is closed. Take $X-U$. Since $X$ has finitely many elements, then so does $X-U$, which implies $X-U$ is closed. Therefore $U$ is open.
A: Let $X$ be a finite topological space with a hausdorff topology. Observe that the topology must be discrete, and consequently, every subset is both, open and closed.
