Just for fun, here's a "third approach". (Although, this one may be slightly over-complicated and/or needs quite a bit of justification.) Since $f$ and $g$ are continuous functions $K\to\mathbb R$ and $K$ is compact, their graphs $\Gamma_f$ and $\Gamma_g$ must be compact. Now, because $f$ is always greater than $g$, $\Gamma_f$ and $\Gamma_g$ are also disjoint. But then $\Gamma_f$ and $\Gamma_g$ must be a positive distance apart, which proves the claim.
Added: to avoid confusion I shall add some further remarks. The graphs $\Gamma_f$ and $\Gamma_f$ are subsets of $K\times\mathbb R$. There are many ways to equip a product of two metric spaces with a product metric. So, $K\times\mathbb R$ is itself a metric space. Since the product metric gives rise to the product topology on $K\times\mathbb R$, the maps $(id_K,f):K\to K\times\mathbb R$ and $(id_K,g):K\to K\times\mathbb R$ are continuous, so their images, which are precisely $\Gamma_f$ and $\Gamma_g$, must are compact. Now, if $X$ is a metric space and $A,B\subseteq X$ it makes sense to define $d(A,B)=\inf\limits_{a\in A, b\in B}d(a,b)$. This usually isn't a metric, but we can prove that if $A$ and $B$ are compact and disjoint, $d(A,B)$ is positive. (Although this last fact is about as hard to prove as the original problem, which is why I am saying that this approach may be slightly over-complicated.)