How to prove $0\leq a_n \leq 3$ for all but finitely many $n$

Question

A sequence $(a_n)$ has $1$ as $\liminf$ and $2$ as $\limsup$. Then it follows:

1. $1\leq a_n \leq 2$ for all $n$

2. $1\leq a_n \leq 2$ for all but finitely many $n$

3. $1\leq a_n \leq 2$ for infinitely many $n$

4. $0\leq a_n \leq 3$ for all but finitely many $n$

Since $\liminf = 1 \neq 2 = \limsup$, so sequence $(a_n)$ does not converge and from given information we say that it is bounded So given sequence has more than one limit point. By definition of limit point, there exist an open set that contains infinitely many values of the sequence. Correct option is 4. How can I prove it and discard the other options.

Answer 1

The correct answer is 4.

$$a_n =\begin{cases} 1- \frac{1}{n} & \text{ for } n \text{ even}\\ 2+ \frac{1}{n} & \text{ for } n \text{ odd}\\ \end{cases}$$

is a counterexample for 1., 2., 3.

4. is not difficult to prove.