Explanation that Twistor Space of $S^4$ is $\mathbb{C}P^3$?

I am attempting to read Atiyah's paper on self-duality in four-dimensional Riemannian geometry, and I came across the following basic example:

Let $S_-$ be the $SU(2)$-bundle of anti-self dual spinors over $S^4$. Then the total space of the projectivised bundle $PS_{-}$ over $S^4$ is $\mathbb{C}P^3$.

I was wondering if someone could give a brief proof/explanation of this fact (after searching the internet I find many references to this fact, but no complete explanations)? I understand that the quaternionic Hopf fibration exhibits $\mathbb{C}P^3$ as an $S^2$-bundle over $S^4$, but I only vaguely see that this bundle is in fact the projectivised ASD spinor bundle. (Basically I have tried to write down transition functions for both for the usual splitting of $S^4 = \mathbb{H}P^1$ into two hemispheres. I was hoping that someone could give a better reason and/or the details of this procedure.)

Thanks in advance, and sorry if this question is basic! Everyone seems to just claim that the total space is obviously $\mathbb{C}P^3$, but to me it is not obvious :(.

(In a similar vein, if someone could describe and explain the twistor space of $\mathbb{C}P^2$, that would also be great!)

I am sure there are many ways to see this but here is a quick one. The oriented frame bundle of $\mathbb S^4$ is $SO(4)\to SO(5)\to \mathbb S^4$. The spin cover of this is $Spin(4)\to Spin(5)\to S^4$ which is the same as $Sp(1)\times Sp(1)\to Sp(2)\to \mathbb S^4$. Up to conjugation there is only one $Sp(1)\times Sp(1)$ in $Sp(2)$ so we can think of this just being the canonical diagonal embedding. The half-spin reps $\Delta_\pm$ of $Spin(4)=Sp(1)\times Sp(1)$ are just the quaternionic multiplications on $\mathbb H=\mathbb R^4$ by the first and the second factor respectively. Therefore the unprojectivized $SU(2)$ bundle you are interested in is just $Sp(1)\to Sp(2)/(Sp(1)\times 1)=S^7\to S^4$ which of course is just the Hopf bundle. Taking the quotient of $S^7$ by $S^1\subset Sp(1)$ gives you $\mathbb{CP}^3$.