Finite Topological spaces there is  a step which is not clear  to me on this link http://www.math.uchicago.edu/~may/MISC/FiniteSpaces.pdf
 page 3 Lemma 2.2  .how we get that   Uy is contained f−1(V) at the last step to show that f is continuous ??? 
 A: I don't know which step you are particularly talking about. But, since it's a very nice topic, let me try to solve your problem.
The lemma you told is related to the observation that the category of finite topological spaces is isomorphic to the category of finite partial ordered sets.
I believe that you asked about the proof of the "converse of the theorem".
Solution: Let $f: X\to Y $ be a order preserving function between partial ordered sets. You construct finite topological spaces from each partial ordered set (I guess you've already known it).  By definition, in the obtained topological space $X$,  $ x\leq y $ if and only if $x\in U_y $, in which $ U_y $ is the smallest open set containing $ y $.
So, you may say that, for $ z\in X $, $U_ z = \left\{ x\in X : x\leq z \right\} $. 
Ok, now we need to prove that " $f$ is continuous ". Well, I guess you know that the "smallest open sets" of a finite topological space forms a base for the topology. 
So, you may consider just these open sets to prove that $ f $ is continuous. Given $x\in X $ and $ U_ {f(x)}\subset Y $ (in which $ U_{f(x)} $ is the smallest open set containing $ f(x) $), you only need to prove that $ U_x\subset f^{-1}(U _{f(x)}) $.
Given $ a\in U_x $, we have that $a\leq x $. And, thus, $f(a)\leq f(x) $, which means that $f(a)\in U _{f(x)} $. So $a\in f^{-1} (U_{f(x)}) $.
And, therefore, $ U_x\subset f^{-1}(U _{f(x)}) $.
