Let $X$ be a topological space and $X^*$ be its supspace. It is stated in my textbook that if $c(A)$ represents the closure of set $A$ in $X$, then $c(A) \bigcap X^*$ is closed in $X^*$.
A closed set is one which contains all its limit points, and a limit point of a set is a point such that every open set containing it contains a different point from the aforementioned set.
Let $l$ is an external limit point of set $A$. If there is an open set containing $l$, it has to contain a point in $A$- let's call it $p$. Let the open set containing $p$ and $l$ not contain any other point in $A$. I don't see why that should be a problem at all. Let the subspace $X^*$ contain $l$, but not $p$.
$c(A) \bigcap X^*$ will contain $l$, but $l$ will be not a limit point of $A$, as there is an open set containing $l$ and no point in $A \bigcap X^*$ ($p$ is not there in $X^*$). How is $A \bigcap X^*$ closed in $X^*$ then?