$\def\bdy{\operatorname{bdy}}\def\interior{\operatorname{int}}$(1) Prove that a function from a metric space X into the metric space Y is continuous if and only if for each $A \subset X , f(\bar A) \subset \overline {f(A)} .$
(2) Prove that if $f$ is a one to one mapping from the metric space X into the metric space Y, show that $f$ is a homeomorphism if and only if, for each $A \subset X, f(\bar A)=\overline {f(A)}.$
Attempt at one. $\bar A=\interior(A) \cup \bdy A$. So x $\in A$ then $f(x) \in f(\bar A)$ also $f(x) \in \overline {f(A)}$. If $ x \in \bdy A$ then $f(x) \in f(\bar A)$ like wise $f(x) \in \overline {f(A)}$. But $f(\bar A)=f(A) \cup f(\bdy A) $ $f(\bar A) $maybe a open set where $\overline {f(A)}$ is a closed set containing the $\bdy f(A) \cup f(A).$ Now $\bdy f(\bar A) \not= \bdy\overline {f(A)}$ because $f(\bar A)$may be an open set in Y. Where as $\overline {f(A)}$ is closed so it contains all boundary points. Do not really have a clue how to do the converse.
Attempt at 2: Since $f$ is one to one and a homeomorphism we know that for every $f(a) \in f( A) $ are distinct so we need only to show that $\bdy f(A)=f(\bdy A).$ Not quite sure how do prove that and also not sure how to prove converse.