Let $K_n:=(n,\infty)$ for $n\in\mathbb{N}.$ Prove that $\bigcap_{n=1}^\infty K_n=\emptyset.$ I did a proof for exercise 2.5.9 of Introduction to Real Analysis: 4th edition by Bartle and Sherbert. The exercise is:

Let $K_n:=(n,\infty)$ for $n\in\mathbb{N}.$ Prove that $\bigcap_{n=1}^\infty K_n=\emptyset.$

My proof was significantly different than the solution manual, so I wanted to see if it's still correct, and if not, is there any way I can improve? Thank you!
Proof: Assume to the contrary that
$$ \bigcap_{n=1}^\infty K_n\neq\emptyset.$$
Then assume $x\in\bigcap_{n=1}^{\infty} K_n.$ Then we have $x>n$ for all $n\in\mathbb{N}$, which implies $x$ is an upper bound of $\mathbb{N}$. However, by the Archimedean Property*, $\mathbb{N}$ is not bounded above, so this is a contradiction.
$\therefore\bigcap_{n=1}^\infty K_n=\emptyset.\quad\blacksquare$
*Note that in the lectures for my Real Analysis class, the Archimedean Property was simply defined as "$\mathbb{N}$ is not bounded above," and we proved this fact in class.

Improvement is a relative entity, just let me offer the direct way to solve the exercise you set out.
You need not pass by contraposition, the direct approach is a more direct path towards $\blacksquare\,$:
Fix an arbitrary $x\in\mathbb R$.
Then choose $n\in\mathbb N$ such that $x\leqslant n$, whence $x\notin K_n=(n,\infty)\,$.
(E.g., $n=\lfloor x\rfloor +1$ does the job.)
Thus, the intersection of all $K_n$ is empty.$\quad\blacksquare$
