Closed set

This is in Lemma 3.1 of Hartshorne's Algebraic Geometry.

"An open subset of an open subset is open" is the trick.

Proposition. Let $Y$ be a topological space. $Z \subset Y$ is an closed set if and only if there exists an open cover $\{ U_\alpha \}$ such that $Z \cap U_\alpha$ is closed in $U_\alpha$ for all $\alpha$.

Proof) 

$(\Rightarrow)$: Suppose $Z$ is closed. $Y$ is an open cover of $Y$ and $Z \cap Y$ is closed in $Y$.

$(\Leftarrow)$: If $Z \cap U_\alpha$ is closed in $U_\alpha$, then $Z^c \cap U_\alpha$ is open in $U_\alpha$. Hence $Z^c \cap U_\alpha = Z_\alpha \cap U_\alpha$ for some $Z_\alpha$ open in $Y$. In particular, $Z^c \cap U_\alpha$ is open in $Y$.

$$Z^c = Z^c \cap Y  = Z^c \cap \left ( \bigcup U_\alpha \right ) = \bigcup ( Z^c \cap U_\alpha)$$

hence $Z^c$ is open. Hence $Z$ is closed.