Question: (James Munkres page 104 Theorem 18.1) Let $X$ and $Y$ be topological spaces; let $f: X \rightarrow Y$. If $f$ is continuous, then for every subset $A$ of $X$, one has $f(\overline{A})\subset \overline{f(A)}$. I can follow the argument in the book. My question is why cannot one has equality here, that is, $f(\overline{A}) = \overline{f(A)}$, please? Is there an easy counter example, please? Thank you!

Answer: 
There is a nice symmetry here: $f$ is continuous iff for all $A \subset X$ we have $f[\overline{A}] \subset \overline{f[A]}$ (which means intuitively that a point "close to" $A$ has an image "close to" $f[A]$), and $f$ is closed (maps closed sets to closed sets) iff for all $A \subset X$ we have $\overline{f[A]} \subset f[\overline{A}]$.

Proof of the latter: suppose $f$ is closed, and $A \subset X$. Then $\overline{A}$ is closed in $X$, so $f[\overline{A}]$ is closed in $Y$, and it contains $f[A]$, so it also contains its closure. For the reverse, suppose we have the inequality for all $A \subset X$, and $C \subset X$ is closed. Then $\overline{f[C]} \subset f[\overline{C}] = f[C] \subset \overline{f[C]}$, showing that $f[C]$ is closed.

So we have equality for all $A$ exactly when $f$ is closed and continuous, so we just have to find non-closed continuous maps. And this is what the other answers provide...
Let $A=\mathbb{Z}\_+\subset \mathbb{R}\_+$ with the standard topology, and let $f(x) = \dfrac{1}{x}$. 

The closure of $A$ is $A$, so the image of $f$ does not change. The closure of $f(A)$ is $\{0\} \cup f(A)$.