Where is my argument that $\int_{-1}^{1} \sqrt{1-x^2}dx=0$ wrong?

For $x \in [-1,0)$, you can't have $$ u = 1-x^2 \implies x = (1-u)^{1/2} $$ thus your change of variable is not valid over $[-1,0)$.

You would better write by parity $$ \int_{-1}^{1} \sqrt{1-x^2}dx=2\int_0^{1} \sqrt{1-x^2}dx $$ then use the given change of variable.

You may only do a u-substitution when it is bijective on your integration domain. You can solve this problem by breaking up your integral into $\int_{-1}^0$ and $\int_0^1,$ and using $x=-(1-u)^{1/2}$on the first integral, and $x=(1-u)^{1/2}$ on the second.