Given a vector $v$ on $\mathbb{R}$. Let $\mathbb{R}_v^2$ be the orthogonal complement to $\mathbb{R}v$. Then $\mathbb{R}_v^2$ has a positive orientation that obeys the right hand rule.
Let $\theta$ be any angle. Then we can take $\mathbb{R}_v^2$ along the axis $\mathbb{R}_v$. A choice of $v$ and $\theta$ then gives all the element in $SO(3)$.
Here I'm not sure why we can have such choice of $v$. Also, I'm not sure how this provides all the elements in $SO(3)$.
Any help will be appreciated!
The key is that you are in an odd-dimensional vector space: Any matrix $A\in SO(3)$ (or to be more general $A\in\mathbb{R}^{(2n+1)\times (2n+1)}$ for all $n\in\mathbb{N}$) has at least one real eigenvalue, since its characteristic polynomial has odd degree (and thus a real zero). More precisely, one of these real eigenvalues is $1$ due to the condition $\det A=1$.
The subspace $\mathbb{R}_v$ spanned by a corresponding eigenvector $v\in\mathbb{R}^3$ is then fixed under $A$, i.e. is invariant under rotation, and so $A$ is completely determined by its action on $\mathbb{R}_v^2$, which is given by a rotation around some angle $\theta\in\mathbb{R}$.
Formally, this means that each matrix in $SO(3)$ is conjugate to a matrix of the form $$\left(\begin{matrix}\cos\theta & -\sin\theta &0\\ \sin\theta&\cos\theta&0\\0&0& 1\end{matrix}\right).$$