Are these implications about topology true or false?

If $f: \mathbb{R}\to\mathbb{R}$ is a continuous function ($\mathbb{R}$ with euclidean distance) are these sentences true or false?

1) $f(A)$ open $\Rightarrow A$ open

2) $A$ open $\Rightarrow f(A)$ open

3) $f(A)$ closed $\Rightarrow A$ closed

4) $A$ closed $\Rightarrow f(A)$ closed

Well, I found that 2) is false, in fact if I take $A = (-1,1)$ and $f(x) = x^2$, then I have $f(A) = [0,1)$, which is not open. 4) Is also false because if I take $A = [0,\infty)$ and $f(x) = \arctan(x)$, then $f(A) = [0,\pi/2)$ is not closed.

I think that 1) and 3) are false. If I'm right, I have to prove it. Can you give me a little help? Thanks in advice.

All are false. For (1) let $f(x)=x^2$ and pick a suitable $A$. Here is a hint. If we tried $A = (1,2)$ then we would have $f(A)=(1,4)$. To make it a counterexample we should change $A$ a little, by adding some more stuff to $A$ while keeping $f(A)$ the same. So set $A = (1,2) \cup \text{[something]}$ where $f(A) = (1,4)$ still, meaning that the $[\text{something}]^2 \subset (1,4)$; and pick the $[\text{something}]$ to be a not open set.

I hope this is a good enough hint for (1). You can do something very similar for (3).

1 is not true. Consider $f(x) = x \sin(x)$ on $x \geq 0$, and $0$ on $x < 0$. Then $f([0,\infty)) = \mathbb{R}$.

The same example works for 3 if you use $(0, \infty)$ instead.

How to come up with this? I immediately thought "this has to be false", for reasons I'm not clear about; so I next tried to find the simplest possible answer. The simplest way would be if $f(A) = \emptyset$ but $A$ were nonempty - impossible. So the next simplest is if $f(A) = \mathbb{R}$ but $A$ is not open (and not closed, for question 3). We need $A$ not to be compact, because the continuous image of a closed bounded set is closed and bounded; the simplest non-compact nonopen sets are $[a, b)$.

Easy: just make $f$ take all values on $[0, \infty)$. How can this be done? Uurgh, this is annoying, I should just be able to draw a picture; but if you really must, something which oscillates ever wider as $x$ gets bigger. $\sin$ does the trick if we blow it up a bit.

More counterexamples:

For (1):

Take $f(x) = \left\lvert x\right\rvert$, $\forall x \in \mathbb{R}$ and $A = \left\{-\frac{3}{2}\right\}\cup\langle1, 2\rangle$. Then $f(A)=\langle1, 2\rangle$, which is open, but $A$ is not open.

For (3):

Take $f(x) = 1$, $\forall x \in \mathbb{R}$ and $A = \langle 0, 1\rangle$. Then $f(A) = \{1\}$, which is closed, but $A$ is not closed.

Note that (1) and (3) are in fact true when $f$ is an injective continuous function: if $f(A)$ is open then $A = f^{-1}\left(f(A)\right)$ which is open as a preimage of an open set. Same when $f(A)$ is closed.