Continuity and adherence For two topological spaces $E$ and $F$ 
Please how to prove that $$f(\overline{A})\subset \overline{f(A)},\forall A\Rightarrow f:E\rightarrow F ~\text{is continuous}$$
Thank you

Suppose that $f$ is not continuous. Then there is some closed set $C\subset F$ such that $f^{-1}(C)$ is not closed. Call $A=f^{-1}(C)$. Then there is some $x\in\overline A$ such that $f(x)\notin C$. That is, $f(\overline A)\not\subset C=\overline C=\overline{f(A)}$.
Arternatively, suppose that $f$ is not continuous at some point $x$. Then, there exists a neighbourhood $V$ of $f(x)$ such that $f(U)$ is not contained in $V$ for every neighbourhood $U$ of $x$.
Call $A=f^{-1}(F-V)$. Since for every neighbourhood $U$ of $x$, $f(U)$ intersects $F-V$, there is some $y\in U$ such that $f(y)\in F-V$. that is, $y\in A$. That means that $x\in \overline A$, so $f(x)\in f(\overline A)$. But $f(x)\in V$, so $f(x)\notin F-V=\overline{F-V}=\overline{f(A)}$.
