Why does $u \perp \text{ker}(A-I)$ imply that $u$ eigenvector of A? Say $A \in \text{SO}(3)\setminus \{I\}$ and $\text{dim }(\text{ker}(A-I)) =2$. Let $ u \in \mathbb{R}^3$ and $u \perp \text{ker}(A-I)$.
Why does this imply that $u$ must be an eigenvector of $A$?
 A: Let $P$ be the kernel of $A - I$, and let $L$ be its orthogonal complement. We're assuming $P$ is a plane in $\mathbb{R}^3$, which means $L$ is a line.
Since $A$ is an orthogonal map, it preserves orthogonality. In particular, $AL$ is orthogonal to $AP$. The plane $P$ is an eigenspace of $A$, so $AP = P$. That means $AL$ is orthogonal to $P$. Equivalently, $AL$ is contained in the orthogonal complement of $P$: in other words, $AL \subset L$.
Finally, pick a vector $u \in L$. From the previous paragraph, we know that $Au \in L$. I'll leave the rest to you.
A: $A$ is normal, and you know $1$ has algebraic multiplicity $\ge 2$, so the characteristic polynomial splits over $\mathbb R$, so there must be an orthonormal basis of eigenvectors of $A$. Knowing $\ker(A-I)$ is 2-dimensional doesn't leave much choice for the last eigenvector.
A: This is impossible. A rotation cannot have eigenvalue 1 with multiplicity 2. Purely logically, anything can be concluded from a false hypothesis. Unless you agree to call the null vector orthogonal to everything?
