I was reading Professor Vakil's FOAG, and in exercise 9.2B there is a statement that :
If $V$ is an open subscheme of a closed subscheme will implies that $V$ is a closed subscheme of an open subscheme. But the converse does not hold. (Professor Vakil's provides an counterexample on Stacks Project)
Therefore the following proof is a fake proof for the converse direction:
Let $V$ be a closed subscheme of some open subscheme $U$, then as a closed subset, it's the intersection of $U$ with some closed subset $K$ in $X$ therefore $V = K\cap U$ then $V$ is open subscheme of $K$ which is again a closed subscheme of $X$.
Where does the proof above go wrong?