Calculate $\displaystyle \iiint_B x\text{ d}V$, let $dV=dx\, dy\, dz$ (in that order) where $B$ is tetrahedron with vertices $(0,0,0),(0,1,0),(0,0,1),(1,0,0)$
Here is a picture:
Lets fix $z$. Then $0\leq z\leq 1$, and we want to consider the projection on the $xy$ plane.
We can see that $0\leq x\leq 1-y$, but what about $y$? $y$ must be a function of $z$ and all I see is that $y$ goes from $0$ to $1$?
Treating the $y-z$ plane (i.e. $x=0$ as the base) The region is bounded above by $x=1-y-z$.
Rather than considering $x-y$ plane as you showed in your attempt, you can consider the $y-z$ plane and you will find that $$0 \leq y \leq 1-z$$ $$0 \leq z \leq 1$$
Hence the corresponding region is
$$\int_0^1 \int_0^{1-z}\int_0^{1-y-z} x\,\,\, dx \, dy \, dz$$
Remark:
If you are willing to swap the order of integration, the integral can also be described as : $$\int_0^1\int_0^{1-y} \int_0^{1-x-y} x\, dz\,dx\,dy$$
Notice the similarity, this is not surprising due to symmetric property of the tetrahedron.