Theorem 17.2 on page 94: Let $Y$ be a subspace of X. Then a set A is closed if and only if it equals to the intersection of a closed set of $X$ with $Y$.
In his proof for one of the directions, he says assume $A$ is closed in $Y$. Then $Y-A$ is open in Y, so that by definition it equals to the intersection of an open set $U$ of $X$ with $Y$. The set $X-U$ is closed in $X$, and $A=Y\cap(X-U)$, so that $A$ equals the intersection of a closed set of $X$ with $Y$, as desired.
Could someone please explain why we have $A=Y\cap(X-U)$? Thanks!
$$Y-A = U \cap Y$$ Taking the relative complement of both sides with respect to $Y$ yields $$A = Y - (U \cap Y) = Y \cap (X - U).$$