I want to prove $SO(3)$ is simple without referring to the proof in- Springer-Verlag Naive Lie theory. Specifically the proof in section 2.3 which talks about how $SU(2)$ is closely related $SO(3)$.
A unit quaternion in the form $q=a+bi+cj+dk$ can be represented by an element in $SU(2)$ of the form $
q=
\begin{bmatrix}
a+id & -b-ic \\
b-ic & a-id
\end{bmatrix}
$ such that $\det(q)=a^2+b^2+c^2+d^2=1$. We can associate a pure unit quaternion of the form $v=\mathbb{R}i+\mathbb{R}j+\mathbb{R}k$ with a vector in $\mathbb{R}^3$. The quaternion product $v\mapsto qv\bar{q}$ is a rotation of $v$ about the axis $(b,c,d)$ through an angle $\theta=2\cos^{-1}(a)$. It can be noted that $SU(2) \mapsto SO(3)$ is a 2:1 homomorphism since $v\mapsto (-q)v(-\bar{q})$ corresponds to the same rotation of 3D space because $(-q)v(-\bar{q})=qv\bar{q}$. Explicitly working out this homomorphism for a unit quaternion $q=w+xi+yj+zk$, the matrix: $$
Q=
\begin{bmatrix}
1-2y^2-2z^2 & 2xy-2zw & 2xz+2yw\\
2xy+2zw & 1-2x^2-2z^2 & 2yz-2xw\\
2xz-2yw &2yz+2xw & 1-2x^2-2y^2
\end{bmatrix}
$$
is a rotation about $(x,y,z)$ by an angle $2\theta$ where $\cos(\theta)=w$.
To prove that $SO(3)$ is simple, start by assuming $H$ is a nontrivial normal subgroup of $SO(3)$ and eventually I hope to run into a contradiction. Since $H$ is normal by assumption, $\forall g \in SO(3) \text{ and } \forall h\in H: ghg^{-1}\in H.$ Equivalently, $\forall g \in SO(3), gHg^{-1}\subseteq H$.
Sorry if this post was a ramble. I am just seeing if anyone in this community knows how to prove this and can give me a direction to go in. Maybe a geometric way of thinking about this? I will update this question and provide a full proof once completed.
Without loss of generality, assume that your normal subgroup $H$ contains a rotation corresponding to the quaternion $h = A + Bi$. Now let $g$ be the quaternion that corresponds to the rotation that rotates the $\vec i$ vector to some other (arbitrary) $\vec v$ vector. We then have that $ghg^{-1} = A+B v$ for arbitrary vector $\vec v$.
Additionally, if $h = \cos(\theta) + i\sin(\theta)$ (where $A$ and $B$ have been replaced with trigonometric functions) then by composing $h$ with $ghg^{-1}$ it is possible to express rotations of any angle up to $\theta$. See the following figure in the article on versors to see for yourself how it would work.