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?

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.
