Compact sets are bounded: shape of the cover matters? To prove a compact sets is bounded, we assume there's a "open ball cover" (each with R=1) that covers the set. And take maximum distance over the center of the balls +2 as the boundary. Why could we assume this without actually knowing the "shape" of the cover? How do we know if such "open ball cover" is able to cover the set presumably? Maybe such "ball-shaped" cover is not able to cover the set.

The function $d(x,y)$, defined on $K \times K \rightarrow \mathbb R$ , is a continuous (uniformly-continuous, actually) Real-valued function, defined on the compact set $K\times K$. A continuous function on a compact set reaches , takes on a (finite) maximum value, say $M$. This value can serve as the diameter of a ball containing $K$. 
And re the issue of covering, by definition of a cover $S$, any element  $x$ in $K$ is in a set $S_k$ of the cover. Then $\cup$ $S_k${$x \in K, s\in S$} contains every element of $K$. 
