Debreu's theorem: Let $X$ be a topological space that is connected separable or second countable. A binary relation relation $\succsim$ on $X$ is complete, transtitive, and continuous if and only if there exists a continuous function $u$ such that $$u(x)\geq u(y) \iff x\succsim y.$$
Assume that $(X, T)$ is a topological space and $X$ is at most of size continuum: $|X|≤2^{\aleph_0}$. Assume that $(X, \preceq)$ is a weak preference ordering. Assume that the preference ordering $\preceq$ is continuous: the sets $\{(x, y) : y\in X, x\preceq y\} , \{(x, y) : y\in X, x\succeq y\}$ are each closed sets $\forall x\in X$ in the product topology $(X, T)\otimes (X, T)$.
Then, define a function mapping each element to the infimum of the equivalence class defined by the weak relation: $f(x) = \inf [x] = \inf \{(x, y) : x\preceq y\}$. Because we know that the latter set is closed, and because we know every element has at least one element equivalent to it (itself), the output of this function must be equivalent to the input: $\forall x\in X, f(x) \sim x$.
There are at most $2^{\aleph_0}$ elements in $X$, and so there are at most that many equivalence classes, and so the image $f(X)$ is at most $2^{\aleph_0}$ in size. Therefore, we can define an injection $g : f(X) \rightarrow \mathbb{R}$ that maps smaller elements in $X$ to smaller elements in $\mathbb{R}$; such that $g(x)<g(y)$ if and only if $f(x)<f(y)$. Therefore, $U(x) = g(f(x))$ is a utility function that represents the preference ordering $(X, \preceq)$.
A proof of Debreu's representation theorem - that a continuous preference ordering is always represented by a utility function on the reals - should, I think, be harder than this. While it's not as general as many of the Debreu results, and it doesn't prove the converse, I have seen proofs for special cases of this result, like preference orderings on $\mathbb{R}^N$, that were substantially more complicated than this. I assume there's probably an error in this logic; where is it?