Compactness and Lipschitz functions

Question

I am very stumped by this question:

Suppose (K, d) is a compact metric space. Let f be any function, f: K $\rightarrow \mathbb{C}$, not necessarily continuous. Prove that for any $\epsilon > 0$, there exists a finite M such that:

|$f(u) - f(v)$| $\leq$ Md($u,v$) + $\epsilon$

for all u,v $\in$ K

Couple things:

1. Does $\epsilon$ matter here? If the above holds for every $\epsilon$, then |$f(u) - f(v)$| $\leq$ Md($u,v$). Then it becomes a matter of proving that f is Lipschitz.
2. My first intuition was applying the mean value theorem, but unfortunately f is not continuous. Is there another way to approach the problem (i.e. using sequential compactness)?