Continuity in a compact metric space. 
Let $(X,d)$ be a compact metric space and let $f, g: X \rightarrow \mathbb{R}$ be continuous such that $$f(x) \neq g(x), \forall x\in X.$$
  Show that there exists an $\epsilon$ such that $$|f(x) - g(x)| \geq \epsilon, \forall x \in X.$$

I'm assuming he means $\epsilon > 0$. Well, suppose to the contrary that for all $\epsilon > 0$, there exists an $x' \in X$ such that $|f(x') - g(x')| < \epsilon.$ Since $f(x')$ and $g(x')$ are fixed values, we must have $f(x') = g(x')$, a contradiction. 
Seems uh... too easy? I didn't even have to use continuity or compactness? So seems wrong? (I'm really sick, so terrible at math this week, but is this right?)

Your flaw is in assuming that the values $f(x')$ and $g(x')$ are fixed; they are not, since $x'$ depends on $\varepsilon$.
To prove the claim you just need to observe that $|f-g|$ is a continuous function (since it is the composition of continuous functions) which is positive, and that it is defined in a compact set, so it attains its minimum there, and... 
