I have already seen proofs of this problem on the site, but none of them match the one I did, therefore I would be glad if someone could indicate where is the mistake here. Thanks in advance.
Problem: Let $K$ be a subset of a metric space $M$, then if $U$ is open, $U\cap K$ is open in $K$
My proof: Take a limit point $x$ of $K$ that is not in $K$, but is in $U$ (in other words $x \in U \cap(\overline{K}-K)$), then suppose that $K\cap U$ is closed, this implies $\overline{K \cap U}=K \cap U$ and then must contain all limit points of $K\cap U$ since $x$ is a limit point of $K$ and is in $U$, it is also a limit point of $K\cap U$ and therefore must be in the closure, which is absurd since it would imply in $x \in K$.