Is the following argument correct?
Prove that there exists a sequence of nested unbounded closed intervals $L_1\supseteq
L_2\supseteq L_3\supseteq\cdots$ with $\bigcap_{n=1}^{\infty}L_n =
\varnothing$. (An unbounded closed interval has the form $[a,\infty) = \{x\in\mathbf{R}:x\ge a\}$).
Solution. Consider the nested intervals $L_1 = [1,\infty),L_2 = [2,\infty),L_3 = [3,\infty)\cdots$, Now assume that we have an $x\in\mathbf{R}$ such that $x\in \bigcap_{n=1}^{\infty}L_n$ and let $k\in\mathbf{N}$, from hypothesis $x\in L_{k} = [k,\infty)$ then $x\ge k$ but $x\neq k$ as that would imply that $x\not\in L_{k+1}$, therefore it must be that $x>k$, then since $k$ was arbitrary it follows that $x$ is an upper bound for $\mathbf{N}$, but this is impossible since we know that the real numbers possess the archimedian property, consequently $\bigcap_{n=1}^{\infty}L_n = \varnothing$.
$\blacksquare$
Best Answer
Your proof is perfectly OK, but if you delete "but $x≠k$ as that would imply that $x∉J_{k+1}$, therefore it must be that $x>k$," from your proof then it reads more fluently.
All you need is an upper bound for $\mathbf{N}$, which is found by $x\ge k$