Theorem 1   Suppose $Y\subseteq X$ is equipped with the subspace topology, and $A\subseteq Y$ . Then $A$ is closed in $Y$ if and only if $A=Y\cap J$ for some closed set $J\subseteq X$ .

Proof. If $A$ is closed in $Y$ , then $Y\setminus A$ is open in $Y$ , and by the definition of the subspace topology, $Y\setminus A = Y\cap U$ for some open $U\subseteq X$ . Using properties of the set difference, we obtain \begin{eqnarray*} A &=& Y\setminus (Y\setminus A) \\ &=& Y\setminus(Y\cap U) \\ &=& Y\setminus U \\ &=& Y\cap U^\complement. \end{eqnarray*}On the other hand, if $A=Y\cap J$ for some closed $J\subseteq X$ , then $Y\setminus A = Y\setminus(Y\cap J) = Y\cap J^\complement$ , and so $Y\setminus A$ is open in $Y$ , and therefore $A$ is closed in $Y$ .

Theorem 2   Suppose $X$ is a topological space, $C\subseteq X$ is a closed set equipped with the subspace topology, and $A\subseteq C$ is closed in $C$ . Then $A$ is closed in $X$ .

Proof. This follows from the previous theorem: since $A$ is closed in $C$ , we have $A = C\cap J$ for some closed set $J\subseteq X$ , and $A$ is closed in $X$ .