Is this sufficient to prove that a multiple of a Lipschitz function is also Lipschitz? If we have a Lipschitz function, $f: S \rightarrow \mathbb{R}$ and $h(x) = 3f(x)$, can I show that $h: S \rightarrow \mathbb{R}$ is also Lipschitz by the following?
Assume $f: S \rightarrow \mathbb{R}$ is Lipschitz. Then, there is $K > 0$ such that for all $x,y \in S$ with $|f(x) – f(y)| < \frac{K}{3}|x – y|$. Then, choose for $|3f(x) – 3f(y)| < 3|f(x) – f(y)| < \frac{3K}{3}|x – y| = K|x – y|$. So $3f(x) = h(x)$ is also Lipschitz.
I feel there is a hole in my logic somewhere but cannot grasp quite where.

It is perfectly fine. Indeed note that the space of Lipschitz functions is a vector space, hence it makes perfectly sense.
