Let $(X,d)$ be a metric space and $Y \subseteq X$. Suppose $F \subseteq X$ is closed, show that $F\cap Y$ is closed in $(Y,d)$. Conversely show that if $F_1 \subseteq Y$ is closed in $(Y,d)$, then $\exists F \subset X$ such that $F$ is closed and $F_1 = F\cap Y$.
Proof:
$[\Longrightarrow]$ We will show that $F\cap Y$ is closed in $(Y,d)$ by proving that $ \forall x \in F\cap Y, B^Y(x,\epsilon)\cap(F\cap Y) \neq \emptyset, \mbox { } \forall \epsilon > 0 $ Let $x \in F\cap Y$, then $x \in F$ and $x \in Y$. Since F is closed in (X,d), $ B^X(x,\epsilon)\cap F \neq \emptyset \mbox { } \forall \epsilon >0. $ Since $Y\subseteq X$ we have that $B^Y(x,\epsilon) = B^X(x,\epsilon)\cap Y$, $\forall \epsilon > 0$, then \begin{align*} B^Y(x,\epsilon)\cap(F\cap Y) &= B^X(x,\epsilon)\cap Y \cap (F\cap Y) \\ & = B^X(x,\epsilon) \cap (F\cap Y) \end{align*} therefore, $ x \in B^X(x,\epsilon) \cap (F\cap Y) \Rightarrow B^X(x,\epsilon) \cap (F\cap Y) \neq \emptyset, \forall \epsilon > 0. $
$[\Longleftarrow]$ Let $F_1$ be a closed set in $(Y,d)$, then $F_1 = \overline{F_1} \subseteq Y$ \begin{align*} \overline{F_1} =& \bigcap\{K \subseteq Y | K \mbox{ is closed and } F_1 \subseteq K\} \\ =& \bigcap\{K \subseteq Y | K \mbox{ is closed and } F_1 \subseteq K\}\cap Y. \end{align*} The set $F$ is clearly closed in $(Y,d)$, because is an arbitrary intersection of closed subsets of Y. We can take $F = \bigcap\{K \subseteq Y | K \mbox{ is closed and } F_1 \subseteq K\}.$ This set is closed in $(X,d)$ because is closed in a subset of $X$.
Am I correct? any comments or suggestions? I do not want to prove this using the complements, I know it will be easier.