Behaviour of ${f}^{-1}$ under continuous map

Question

Let $f : \Bbb R \to \Bbb R$ be a continuous function. Which of the following is/are always true ?

1. ${f}^{-1}(A)$ is open for all open sets $A \subseteq \Bbb R$

2. ${f}^{-1}(A)$ is closed for all closed sets $A \subseteq \Bbb R$

3. ${f}^{-1}(A)$ is compact for all compact sets $A \subseteq \Bbb R$

4. ${f}^{-1}(A)$ is connected for all connected sets $A \subseteq \Bbb R$

My Attempt:

We know that inverse image of open/closed sets go to open/closed sets under continuous map. So first two options are always true. Also compact set is closed and bounded. So I take $f : \Bbb R \to [-1 , 1]$ which is continuous function, where $f(x) = sinx$ but ${f}^{-1}[-1 , 1]$ is not bounded. So option 3 is not always true. I'm not sure. I have to general idea to prove or disprove option 4. Please clear my doubt in 3,4 options by giving more examples. Thanks in advance.

Answer 1

To disprove (4), your function , $f(x)=\sin x$ is a counterexample. $\{0\}$ is a connected subset of $\Bbb R$. But $$f^{-1}(\{0\})=\{x: \sin x=0\}=\{k \pi: k \in \Bbb Z\}$$ is a countable set in $\Bbb R$, so disconnected!

PS: This is also a counterexample for (3) !