The statement of the lemma is - Let $f:X\longrightarrow Y$ be a quasi-compact morphism of schemes. Then the subset $f(X)$ of $Y$ is closed if and only if it is stable under specialization.
One way is clear, i.e., if $f(X)$ is a closed subset, then it is stable under specialization. Now, to prove the converse, Hartshorne says, we may assume both $X$ and $Y$ to be reduced. Why is that?
I think we need to apply a result which says - If $f:Z\longrightarrow X$ is a morphism of schemes with $Z$ reduced, then $f$ factors through $cl(f(Z))$ with reduced induced structure.
But I don't know how to apply it in this case. Any help will be greatly appreciated.
Passing to underlying reduced subschemes is a functor on schemes, which doesn't change the underlying topological spaces. Both the properties of closedness and of being closed under specialization are purely topological properties, so there is no problem with passing to underlying reduced subschemes.