Write those inequalities out:
x \ge 0, y \ge 0 , z \ge 0, x+y+z \le 1
It is clear that the region $D$ is contained within unit box $0 \le x \le 1, 0 \le y \le 1, 0 \le z \le 1$, because if any of $x$,$y$ or $z$ exceeds one, $x+y+z \le 1$ will be violated even if the remaining coordinates are zero.
Now, let's do it step-by-step. Assuming $x$ and $y$ are fixed. Variable $z$ can range, within confines of $D$ from $0$ to $1-x-y$, i.e. $0 \le z \le 1-x-y$.
Assuming $x$ is fixed, variable $y$ satisfies $0 \le y \le 1-x$.
And we have already established that $ 0 \le x \le 1$.
Here is a visualization using Mathematica: