Suppose $K$ is a compact metric space and $\varepsilon$ is a positive real number. Show that there is a positive integer $N$ such that every set of $N$ points of $K$ includes at least two points $<\varepsilon$ distance apart.
My attempt: Since $K$ is compact, then $K$ has a finite cover $\{N_\frac{\varepsilon}{2}(p)\}_{p\in K}.$ Take $N>\#\{N_\frac{\varepsilon}{2}(p)\}_{p\in K}.$ Let $S\in K$ be a set with $\#S=N.$ By the pigeonhole principle, there is a $R\in\{N_\frac{\varepsilon}{2}(p)\}_{p\in K}$ that contains at least two points of $S$. By construction, the distance between these points is less than $\varepsilon$ distance apart.
Is my proof sound? Am I allowed to assume that $K$ has a finite cover $\{N_\frac{\varepsilon}{2}(p)\}_{p\in K}?$
Your idea seems nice to me.
Note that a compact set is $\textbf{totally bounded},$ i.e, for every $\epsilon>0$ exist $x_1,x_2....x_m$ such that $K \subset \bigcup_{i=1}^mN_{\epsilon}(x_i)$
So you used this property of compacts sets with $\frac{\epsilon}{2}>0$