Consider this example:
$$
3-\frac12,\quad 5+\frac13,\quad 3-\frac14,\quad 5+\frac15,\quad 3-\frac16,\quad 5+\frac17,\quad 3-\frac18,\quad 5+\frac19,\quad\ldots\ldots
$$
It alternates between something approaching $3$ from below and something approaching $5$ from above. The lim inf is $3$ and the lim sup is $5$.
The inf of the whole sequence is $3-\frac12$.
If you throw away the first term or the first two terms, the inf of what's left is $3-\frac14$.
If you throw away all the terms up to that one and the one after it, the inf of what's left is $3-\frac16$.
If you throw away all the terms up to that one and the one after it, the inf of what's left is $3-\frac18$.
If you throw away all the terms up to that one and the one after it, the inf of what's left is $3-\frac1{10}$.
. . . and so on. You see that these infs are getting bigger.
If you look at the sequence of infs, their sup is $3$.
Thus the lim inf is the sup of the sequence of infs of all tail-ends of the sequence. In mathematical notation,
$$
\begin{align}
\liminf_{n\to\infty} a_n & = \sup_{n=1,2,3,\ldots} \inf_{m=n,n+1,n+2,\ldots} a_m \\[12pt]
& = \sup_{n=1,2,3,\ldots} \inf\left\{ a_n, a_{n+1}, a_{n+2}, a_{n+3},\ldots \right\} \\[12pt]
& = \sup\left\{ \inf\left\{ a_n, a_{n+1}, a_{n+2}, a_{n+3},\ldots \right\} : n=1,2,3,\ldots \right\} \\[12pt]
& = \sup\left\{ \inf\{ a_m : m\ge n\} : n=1,2,3,\ldots \right\}.
\end{align}
$$
Just as the lim inf is a sup of infs, so the lim sup is an inf of sups.
One can also say that $L=\liminf\limits_{n\to\infty} a_n$ precisely if for all $\varepsilon>0$, no matter how small, there exists an index $N$ so large that for all $n\ge N$, $a_n>L-\varepsilon$, and $L$ is the largest number for which this holds.
Hint for (a): both sequences are monotone and bounded.
Hint for (b): you can explicitely find $\bar b_n$ and $\underline b_n$. Your intuition is correct, btw.
Hint for (c): $\bar b_n\ge\underline b_n$ always.
Hint for (d): $\bar b_n\ge a_n\ge\underline b_n$ to prove in one direction and use the fact that a convergent sequence satisfies the Cauchy criterion to prove in another direction.
Anothe way to understand $\limsup$ and $\liminf$ is to take all convergent subsequences; then find $\sup$ and $\inf$ of the set of their limits. This approach, for example, instantly gives the answer to (b).
Best Answer
By intermediate value theorem, since $f$ is increasing, continuous and $A$ compact, $$\sup f(A)=f(\sup A).$$ The continuity allowed you to conclude. It's easy to adapt the proof to a set $\{x_n\}_{n\in\mathbb N}\subset A$.