My apologies if this is a duplicate. I am relatively new to participating on here and I could not find this question. That being said, this is an extra problem for an introductory real analysis class I am currently taking, and I would like some critiques on this proof and where I can improve things, as the textbook had no solution or hints. Thanks in advance!
$\text{Show that if}$ $(I_n)_n\in N$ is a set of nested downward sequences of closed intervals and $\alpha = \sup\{A_n: n\in N\}$ $\text{and let}$ $\beta = \inf\{B_n:n\in N\}$, $\text{then}\,$ $[\alpha , \beta] = \bigcap_{n=1}^\infty I_n $.
$\text{Proof}$
First we will show that $[\alpha , \beta] \subseteq \bigcap_{n=1}^\infty I_n$
$\text{Let}$ $A = \{A_n:n\in N\}$ $\text{and}$ let $B = \{B_n:n\in N\}$.
$\text{Then let}$ $\alpha = \sup\{A_n: n\in N\}$ $\text{and let}$ $\beta = \inf\{B_n:n\in N\}$
Let $x \in [\alpha, \beta]$.
Then $\alpha \le x\le\beta$.
Because $A_n$ $\le$ $\alpha$ and $\beta$ $\le$ $B_n$ , then $A_n \le \alpha \le x \le \beta \le B_n$.
Therefore $x \in [A_n, B_n]$ and thus $x \in\bigcap_{n=1}^\infty I_n$.
So $[\alpha, \beta]\subseteq\bigcap_{n=1}^\infty I_n$.
Now we will show that $\bigcap_{n=1}^\infty I_n\subseteq [\alpha, \beta]$.
Let $x \in\bigcap_{n=1}^\infty I_n$.
So $A_n \le x \le B_n$.
Because $\alpha = \sup\{A_n\}$ and $\beta = \inf\{B_n\}$, $\,$ $A_n \le \alpha \le \beta \le B_n$, $\forall \text{n}\in N$.
Suppose that $A_n \le x \le \alpha$. Because $\alpha$ is the supremum of $A_n$, $\exists$ an element $A_x$ belonging to $[An: n \in N]$ such that $x \le A_x$, and so $x \notin [A_x, B_x]$ and therefore $x \notin \bigcap_{n=1}^\infty I_n$, which is a contradiction.
Then suppose that $\beta \le x \le B_n$. Because $\beta$ is the infimum of $B_n$, $\exists$ an element $B_x$ belonging to $[B_n: n \in N]$ such that $B_x \le x$, and so $x \notin$ $[A_x, B_x]$ and therefore $x \notin \bigcap_{n=1}^\infty I_n$, which is a contradiction.
Thus we can conclude that $\bigcap_{n=1}^\infty I_n \subseteq [\alpha, \beta]$.