The same idea that you use to show that $tr$ is zero on finite-rank operators, can be used to see that it is zero on any projection.
Consider a projection $q$ with infinite dimension and codimension. Consider an orthonormal basis $\{\xi_n\}$ such that $\{\xi_{2n}\}$ spans the range of $1-q$. Let $\{e_{kj}\}$ be the matrix units associated with our orthonormal basis, and let $ x=\sum_je_{j,2j}. $ Then $ x^*x=\sum_je_{2j,2j}=1-q,\ \ xx^*=\sum_je_{j,j}=1. $ So $tr(1)=tr(xx^*)=tr(x^*x)=tr(1-q)$. Then $tr(q)=0$. As $q$ was any projection with infinite dimension and co-dimension, we also have $tr(1-q)=0$. Thus $ tr(1)=tr(q+(1-q))=tr(q)+tr(1-q)=0+0=0. $ Now if $p$ is a projection with finite co-dimension, then $tr(1-p)=0$ (as $1-p$ is finite-rank), but then $tr(p)=0$ since $tr(1)=0$.
We have shown that $tr(p)=0$ for all projections in $B(H)$. It is well-known (see here for references) that every operator in $B(H)$ is a finite linear combination of projections. So $tr=0$.
(in part iii one can avoid this last argument: the fact that $tr$ is positive allows one to use Cauchy-Schwarz, and this together with $tr(1)=0$ can be used to see that $tr=0$: indeed, in that case for any $x$ $ |tr(x)|=|tr(1x)|\leq tr(1)^{1/2}tr(x^*x)^{1/2}=0. $ so $tr=0$)