The proof is from Rordam' s book.
If $F$ is a projection in $K$, where $K$ is the $C^*$- algebra of all compact operators on a separable infinite dimensional Hilber space.
Suppose $F$ is infinite, then $F\sim F_0<F$,which implies that $Tr(F-F_0)=0,$ and this entails $F=F_0$.
I don't understand the last step. If $Tr(F-F_0)=0$,we only have $[F]_0=[F_0]_0.$
This doesn't really answer your question per se, since I am unsure what you mean by the trace on the compacts, but here is a proof that the compacts are finite that I've concocted.
The argument given by yourself and completed by @Epsilon works whenever there is a faithful trace - for example, UHF algebras are finite by this argument.
As mentioned, the compacts have a densely defined trace (I was being pretty dense myself earlier). Since projections in $\mathbb{K}$ are finite rank, the trace argument works (this essentially boils down to the case of proving that $M_n(\mathbb{C})$ is finite).
Now to this specific case.
Suppose that $p \sim q < p$ in $\mathbb{K} = \mathcal{K}(\mathcal{H})$, where $\mathcal{H}$ is separable and infinite dimensional. It is clear that $q < p$ means that $q(\mathcal{H}) \subseteq p(\mathcal{H})$, so that $q(\mathcal{H})$ is a subspace of $p(\mathcal{H})$. What do we know about compact projections? Well the image of the closed unit ball is compact, which is true if and only if the image of said projection is finite dimensional. Thus $p(\mathcal{H})$ and $q(\mathcal{H})$ are finite-dimensional, and if we can show that they have the same dimension, then $p(\mathcal{H}) = q(\mathcal{H})$ and hence $p = q$ (why?).
Now Murray von-Neumann equivalence of projections in $\mathbb{K} \subseteq \mathcal{B}(\mathcal{H})$ implies that the range of the projections have the same dimension (why?), so it follows that $p(\mathcal{H}) = q(\mathcal{H})$ since $p \sim q$. Now this implies $p = q$, which contradicts the fact that $q < p$.