My working:
(a) We can see that $[a_n, b_n] \subset [a_{n-1}, b_{n-1}] \subset .... \subset [a_o, b_o]$. So we have nested intervals here which are closed and bounded. Using the theorem of nested intervals, we know that $ \bigcap_{n=0}^{\infty} [a_n, b_n] = [a,b]$, where $a = \sup(a_n)$ and $b = \inf(b_n)$. However, we can also see that $\lim_{n \to \infty} (b_n - a_n) = 0 \implies b=a$. Therefore, $\bigcap_{n=0}^{\infty} [a_n, b_n]$ = {$a$} $\implies a_n \leq a \leq b_n, \forall n \in \mathbb{N}$, where a is our $\xi$.
(b) We have constructed nested intervals in such a way that $[a_n, b_n] \bigcap S \neq \emptyset$, $\forall n \in \mathbb{N}$. This implies that there exists $s_o \in S$ such that $ a_n \leq s_o \leq b_n, \forall n \in \mathbb{N}$ $\implies a_n \leq s_o \leq \sup(S) \leq b_n, \forall n \in \mathbb{N}$ by the completeness axiom. Now suppose that $s_o < \sup(S)$. This would imply that there exists $s_1 \in S$ such that $s_o < s_1 < \sup(S)$ $\implies a_n \leq s_o < s_1 < \sup(S) \leq b_n, \forall n \in \mathbb{N}$ which is a contradiction to part (a), since we cannot have more than one element in the intersection. Therefore, we have $s_o = a = \xi = \sup(S)$.
This is because $s_o = \sup(S)$ and we have $a_n \leq s_o \leq b_n, \forall n \in \mathbb{N}$ and also $a_n \leq a \leq b_n, \forall n \in \mathbb{N}$. By comparing these two inequalities, we can see that $\sup(S) = s_o = a = \xi$.
I think most of the proof is informal, please help me in making a formal proof. Thank you and do correct any mistakes! :)