 Solution

Assume every continuous function $E\to \mathbb{R}$ is bounded. The Euclidean norm function $x\mapsto \Vert x\Vert$ 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 $\Vert x_j-a\Vert <1/j$. The function $f(x)=1/\Vert x-a\Vert$ 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 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.