Prove $f$ is continuous if $f(\overline{A}) \subset \overline{f(A)}$ for all subsets $A$. Let $X$ and $Y$ be topological spaces. Suppose that $f: X \rightarrow Y$ is a function satisfying: For every $A \subset X$, $f(\overline{A}) \subset \overline{f(A)}$. 
Prove $f$ is continuous.
$\textbf{Proof:}$ Let $X$ and $Y$ be topological spaces. Suppose $f: X \rightarrow Y$ is a function satisfying $\forall A \subset X$, $f(\overline{A}) \subset \overline{f(A)}$. 
Let $a \in \overline{A}$ and then $f(a) \in \overline{f(a)}$. 
Let $V \subset Y$ be open with $f(a) \in V$. 
By way of contradiction, suppose $\forall \, \text{open} \, U \subset X$ with $a \in U$, $f(U) \not \subset V$. 
So, $\forall a \in U$, $b \in X \backslash f^{-1}(V)$ is also $b \in U$. 
Then, $a \in \overline{X \backslash f^{-1}(V)} \subset \overline{Y \backslash V}$ $\rightarrow \leftarrow$ (contradiction!) because $V \cap Y \backslash V = \emptyset$. 
$\therefore f$ is continuous. $\,\, \blacksquare$
 A: Your idea basically amounts to showing that $f$ is continuous at each point of $X$, but to carry it out properly, you really should start with an arbitrary $x\in X$. Now you need an $A\subseteq X$ such that $x\in\operatorname{cl}A$. That’s easy: let $A=\operatorname{cl}\{x\}$. (Of course if $X$ is $T_1$, $A$ will just be the singleton $\{x\}$, but we don’t really care what it is as long as it contains $x$.)
Now suppose that $f$ is not continuous at $x$. Then there is an open nbhd $U$ of $f(x)$ such that $f[V]\nsubseteq U$ for every open nbhd $V$ of $x$. Thus, for each open nbhd $V$ of $x$ there is a point $x_V\in V$ such that $f(x_V)\notin U$. Let $S=f^{-1}[Y\setminus U]$. Then for each open nbhd $V$ of $x$ we have $x_V\in V\cap S$, so $x\in\operatorname{cl}S$, and therefore
$$f(x)\in\operatorname{cl}f[S]\subseteq Y\setminus U\,,$$
which is absurd, since $f(x)\in U$. Thus, $f$ is continuous at $x$ and therefore is continuous.
And when we get to this point, we see that it was really $S$ that we wanted: we never actually used $A$ at all.
