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?

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.
