= Solution
The <empty set> is <bounded>. Otherwise choose $p$ in the <compact metric space> $X$. The <open balls> $B(p,n)$, $n=1,2,\ldots$, form an <open cover>. A <finite subcover> exists by <compactness>; because the balls are nested, the largest one already covers $X$. Hence $X\subseteq B(p,N)$ for some $N$.
For $x,y\in X$, the <triangle inequality> gives $d(x,y)\leq d(x,p)+d(p,y)<2N$. Thus
$$
\boxed{\operatorname{diam}X\leq2N<\infty,\quad\text{so }X\text{ is bounded}.}
$$
The proof uses only <compactness> and the <metric>, not a coordinate system.
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