Let $A$ be a non-empty compact subset of a metric space $(X,d)$,then there exist point $u,v\in A $ such that $d(u,v)=\sup\{d(x,y):x,y\in A\}$.
I have no clue how to approach this using the definition of covers for $A$. Any help would be heartly welcome .
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Then function $G: A \times A \to \Bbb{R}$ such that $G(x,y)=d(x,y)$ is continuous with respect to the product metric: $$D((x_1,y_1),(x_2,y_2)=\sqrt{d(x_1,x_2)^2+d(y_1,y_2)^2}$$
Since $A \times A$ is compact on the product metric space,then $G(A)$ is a compact subset of $\Bbb{R}$ and thus it has a maximum element of the form $d(x_0,y_0)$
Choose sequences $\{x_n\}$ and $\{y_n\}$ such that $d(x_n,y_n)\to \sup\{d(x,y):x,y\in A\}$. This is possible as we can do, $\sup\{d(x,y):x,y\in A\}\geq d(x_n,y_n)\geq \sup\{d(x,y):x,y\in A\}-\frac{1}{n}$. Since $A$ is compact we have a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ converging to some $u$. Then the sequence $\{y_{n_k}\}$ has a convergent subsequence $\{y_{n_{k_l}}\}$ coverging to some $v$. Then, $\{x_{n_{k_l}}\}$ also converges to $u$ as it is a subsequence of the convergent sequnce $\{x_{n_k}\}$. Hence, $$0\leq\sup\{d(x,y):x,y\in A\}- d(u,v)$$$$\leq \sup\{d(x,y):x,y\in A\}-d(u,x_{n_{k_l}})-d(x_{n_{k_l}},y_{n_{k_l}})-d(v,y_{n_{k_l}})$$$$\leq\frac{1}{n_{k_l}}-d(u,x_{n_{k_l}})-d(v,y_{n_{k_l}})\to 0\text{ as }l\to\infty$$$$\implies d(u,v)= \sup\{d(x,y):x,y\in A\}.$$
Here I used the fact, for a metric space compactness implies sequentially compactness. Actually if $X$ is a compact metric space then for for any sequence $\{x_n\}$ in $X$ we have a convergent subsequence $\{x_{n_k}\}$. To prove this, it is enough to show, every countably infinite set has a limit point in $X$ as any sequnce with finite range has always a converget subsequence. So if poosible let $Y=\{y_n\}$ be a countably infinte subset of $X$ without any limit point in $X$. Then, $Y_k:=\{y_n:n\geq k\}$ is closed set in $X$ for each $k$. Hence $\{X\backslash Y_k\}$ is an open cover of $X$ having no finte subcover. So we are done.