I know this is very similar to the base rudin question $(1/n)\cup {0}$ but I’m not sure my adaptation of that solution is accurate.
Consider an open cover G$_{\alpha}$ of K. Let 1 be in G$_{\alpha_0}$.
G$_{\alpha0}$ is open, so there exists $r \gt 0stB(1,r) \subset G_{\alpha_0}$.
By the Archimedean property there exists an integer $N st (1/N)\lt r$. So for $n\geq N, (1-(1/n))\in B(1,r)\subset G_{\alpha_0}$.
Suppose $(1-(1/i))\in G_{\alpha_i}$ for $i=1,2,....N$.
Then, we have a finite subcover $\cup_{i=1}^N G_{\alpha_i}$ $\cup$ G$_{\alpha0}$ and K is compact by the definition.
I’m mostly stuck on the Archimedean step, I don’t intuitively see that part and I’m sort of just guessing what needs to change to make the solution of ${1/n} \cup {0}$ apply to this particular problem
P.S. I tried REALLY hard to use the math Jax tutorial this time!
What you did is correct. The Archimedian property of the reals assures us that there is some natural $N$ such that $\frac1N<r$. But then, if $n\geqslant N$, we also have $\frac1n<r$ and therefore$$\left\lvert\left(1-\frac1n\right)-1\right\rvert=\frac1n<r;$$in other words, $1-\frac1n\in B(1,r)$;