Why did this extraneous root creep into the solution?

When $x=\frac12$, you have $\arcsin(x)=\frac{\pi}{6}$ and $\arcsin(1-x)=\frac{\pi}{6}$

so $\sin (\arcsin (1-x))=\sin \left( \frac {π}{6} \right)$ and $\sin \left( \frac {π}{2} + 2\arcsin (x)\right) =\sin \left( \frac {5π}{6} \right) $

showing your third line would be a correct equality when $x=\frac12$ since $\sin \left( \frac {π}{6} \right)=\sin \left( \frac {5π}{6} \right) =\frac12$

but your second line would not be a correct equality when $x=\frac12$ since $\frac {π}{6}\not = \frac {5π}{6}$

and it is this use of $\sin$ which creates an equality which was not in the original expression.

$\arcsin (1-x) - 2\arcsin (x) = \frac{π}{2} \Rightarrow x=0 \text{ or } x=\frac12$ is a correct statement, but checking shows $x=0 \Rightarrow \arcsin (1-x) - 2\arcsin (x) = \frac{π}{2}$ while $x=\frac12 \not \Rightarrow \arcsin (1-x) - 2\arcsin (x) = \frac{π}{2}$

$$-2\arcsin(x)=\dfrac\pi2-\arcsin(1-x)=\arccos(1-x)$$

Now using Principal values of $\arccos$,

$0\le-2\arcsin(x)\le\pi\iff0\ge\arcsin(x)\ge-\dfrac\pi2\implies-1\le x\le0$

Now let $\arcsin x=y\implies x=\sin y$

and $\arcsin(1-\sin y)=\dfrac\pi2+2y$

$$1-\sin y=\sin\left(\dfrac\pi2+2y\right)=\cos2y=1-2\sin^2y$$

$$\implies \sin y=0,\dfrac12$$ but $\sin y=x\le0$

The sine function is not one-to-one. When you applied that function you also got

$\sin(\arcsin(1-x))=\sin(\frac{\pi}{2}-2\arcsin(x))$,

as this RHS equals the one you intended. Note the sign change on the right side.

Then $x=\frac{1}{2}$ satisfies

$\arcsin(1-x)+2\arcsin(x)=\frac{\pi}{2}$.