If f is continuous does this means that the image under f of any open set is open? If $(X,d_X)$ and $(Y,d_Y)$ are metric spaces, and $f : X → Y$ is a continuous map, is it true that for any open set $U ⊂ X$, the set $f(U)$ is open in Y ?
 A: No.  The easiest counter example is $f(x) = c$.  Then for all $U$, $f(U) = \{c\}$ is not open. 
In general if $f$ is a real function, not injective,  then open intervals can be found around a local max or min, the image of which will be an (at least) half-closed interval.  Example: $f(x) = x^2$ has a minimum at $x = 0$ so for $a < 0 < b$, $f((a,b)) = [0, \max(a^2, b^2))$ is not open.
There are many more counter examples.
A: No, but if you are looking for an interesting paper in this direction, consult:

Velleman, D. J. (1997). Characterizing continuity. American Mathematical Monthly, 318-322.

You can find a link to the article on JSTOR here. 
A preview of the start:

A: This is not true in general.
Let $f: \Bbb R \to \Bbb R$ be defined as the constant function $f(x) = 2$.  Then $f$ is continuous, but every open set is mapped to $\{2\}$, which is not open.
A: Another counter-example is for $U=(0,4 \pi)$ we have $\sin U = [-1,1]$
A: Each function $f:(X,\tau_1)\to (X,\tau_2)$ where $\tau_1$ is Discrete Topology, is continous but there is no guarantee that $f$ be open
