Do discontinuous preferences imply no continuous utility function?
The easiest way to prove it is using the 'old' definition of continuity. 
$\succ$ is continuous iff whenever $x\succ y$, there exists neighborhoods of $x$ and $y$, $B_x, B_y$, such that all $z\in B(x)$ and $z'\in B(y)$, $z\succ z'$.

Suppose $x\succ y$. Because $u$ represents $\succ$, $u(x)&gt;u(y)$. Let $2\epsilon=u(x)-u(y)$. Because $u$ is continuous, there exists some $\delta&gt;0$ such that for all $z\in B_{\delta}(x)$, $u(z)&gt;u(x)-\epsilon$. Similarly, for all $z'\in B_{\delta}(y)$, $u(z')&gt;u(y)+\epsilon$. But then for all $z\in B_{\delta}(x)$ and $z'\in B_{\delta}(y)$, $z\succ z'$ as required.