Prove $(K_n)_{n\in N}$ is a nested sequence of nonempty closed and bounded subsets of $\mathbb{R}$.

For the following problem below, did I present a reasonable argument for why closedness of the interval cannot be dropped? Thank you

Prove if $$(K_n)_{n\in N}$$ is a nested sequence of nonempty closed and bounded subsets of $$\mathbb{R}$$ then $$\cap_{n\in\mathbb{N}} K_n$$ is nonempty. Show that, if the requirement of closed is omitted, then $$\cap_{n\in\mathbb{N}} K_n$$ can be empty.

$$\textbf{Solution:}$$ Suppose $$K_{n+1} \subseteq K_n$$ with $$K_n = [a_n,b_n]$$; $$n\in \mathbb{N}$$. So, observe that $$[a_{n+1}, b_{n+1}] \subseteq [a_n,b_n]$$ for all $$n\ge 1$$. So, $$a_n \le a_{n+1} \le b_{n+1} \le b_n$$ for all $$n\ge 1$$. So, $$a_n \le a_{n+1}$$ for all $$n\ge 1$$ and $$b_{n+1} \le b_n$$ for all $$n\ge 1$$. Therefore, $$\{a_n\}$$ and $$\{b_n\}$$ are non-decreasing and non-increasing sequence respectively. Since $$K_{n+1} \le K_n$$ for all $$n\ge 1$$, then $$K_1 \ge K_2 \ge K_3 \ge K_4 \ge \dots \ge K_n \ge K_{n+1} \ge \dots.$$ Thus, $$a_n, b_n \in K_1$$, implying $$a_n, b_n \in [a_1,b_1]$$ for all $$n\ge 1$$. Hence, the sequence $$\{a_n\}$$ and $$\{b_n\}$$ are bounded. Thus, $$\{a_n\}$$ is a non-decreasing sequence bounded above by $$b_1$$, then the sequence $$\{a_n\}$$ must converge to its least upper bound, call it $$x.$$ Therefore, $$\lim_{n\to\infty} a_n = x$$ where x is the least upper bound of $$\{a_n\}$$, implying $$a_n \le x$$ [*].

Furthermore, $$\{b_n\}$$ is a non-increasing sequence bounded below by $$a_1$$, then $$\{b_n\}$$ must converge to its greatest lower bound, call it $$y$$. Therefore, $$\lim_{n\to\infty} b_n = y$$ where $$y$$ is greatest lower bound of $$\{b_n\}$$, and $$y\le b_n$$ [**]\$.

Moreover, observe that $$a_n \le b_m$$ for all $$m,n \in \mathbb{N}$$, implying the least upper bound $$\{a_n: n\in \mathbb{N}\} \le$$ greatest lower bound$$\{b_m : n\in \mathbb{N}\}$$, implying $$x\le y$$ $$[***]$$, from $$[*]$$, $$[**]$$, and [***], we arrive at $$a_n \le x\le y\le b_n$$ for all $$n\ge 1$$ [A]. So $$[x, y] \subset [a_n , b_n]$$ for all $$n\ge 1$$ implying $$\displaystyle{[x,y] \subset \cap_{n=1}^\infty K_n} [****]$$. Thus, $$\cap_{n=1}^\infty K_n \ne \emptyset.$$ Equivalently, we can say $$\cap_{n=1}^\infty K_n$$ is non-empty.

Now, let us consider the case in which closedness of interval is dropped. Let $$K_n = (0, \frac{1}{n}); n\in \mathbb{N}$$. Then, $$K_{n+1} \subset K_n$$, implies $$(K_n)_{n\in\mathbb{N}}$$ is nested sequence. Moreover, $$K_n = (0, \frac{1}{n})$$ is bounded for each $$n$$. However, $$\displaystyle{\cap_{n=1}^\infty (0, \frac{1}{n}) = \emptyset}$$, implying $$\displaystyle{\cap_{n=1}^\infty K_n = \emptyset.}$$ Hence, $$\displaystyle{\cap_{n=1}^\infty K_n}$$ is empty with $$K_n = (0, \frac{1}{n}).$$ Hence closedness of the interval cannot be dropped.

Your solution is wrong from the start, since it assumes that the only closed and bounded subsets of $$\mathbb R$$ are the intervals $$[a,b]$$, with $$a\leqslant b$$.
You can prove the statement that you want to prove in several ways. One of them is to define $$a_n=\min K_n$$ and then to prove that the sequence $$(a_n)_{n\in\mathbb N}$$ must converge and that its limit must belong to every $$K_n$$.
You might want to justify why $$\bigcap_n (0, \frac1n) = \emptyset$$ some more:
If $$x$$ is in the intersection, this means that $$x>0$$ but then there is some $$m \in \Bbb N$$ such that $$x < \frac{1}{m}$$ (by the Archimedian property of $$\Bbb R$$, e.g.) and then for this $$m$$, $$x \notin (0,\frac1m)$$ and so $$x \notin \bigcap_n (0, \frac1n)$$, contradiction.