= Solution
Assume every <continuous function> $E\to\mathbb R$ is bounded. The <Euclidean norm> function $x\mapsto\|x\|$ is continuous, so its boundedness proves that $E$ is a <bounded set>.
If $E$ were not a <closed set>, there would be $a\in\overline E\setminus E$. Choose $x_j\in E$ with $\|x_j-a\|<1/j$. The function $f(x)=1/\|x-a\|$ is continuous everywhere on $E$, because its denominator is nonzero there. Yet $f(x_j)>j$, contradicting the hypothesis. Thus $E$ is closed. Together with the two previous implications this proves \b[all three conditions are equivalent]. The empty set satisfies all three conditions trivially. This is the Euclidean case of the <bounded-continuous-function characterization of compact metric spaces>.
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