As a part of self study, I am trying to prove the following statement:

Suppose $X$ and $Y$ are topological spaces and $f: X \rightarrow Y$ is a map. Then $f$ is continuous if and only if $f(\overline{A})\subseteq \overline{f(A)}$, where $\overline{A}$ denotes the closure of an arbitrary set $A$. 

Assuming $f$ is continuous, the result is almost immediate. Perhaps I am missing something obvious, but I have not been able to make progress on the other direction. Could anyone give me a hint which might illuminate the problem for me? 

Because the property is stated in terms of closures, it's I think slightly easier to use the inverse image of closed is closed equivalence of continuity instead:

If $C$ is closed in $Y$, we need to show that $D = f^{-1}[C]$ is closed in $X$.

Now using our closure property for $D$: $$f[\overline{D}] \subseteq \overline{f[D]} = \overline{f[f^{-1}[C]]} \subseteq \overline{C} = C,$$ as $C$ is closed.

This means that $\overline{D} \subseteq f^{-1}[C] = D$, making $D$ closed, as required.
