Commutant of bounded linear operators on a Hilbert space

If $A$ is in the commutant and $x$ is any non-zero vector, then the fact that $A$ commutes with the orthogonal projection to the line spanned by $x$ implies that $Ax$ is a scalar multiple $\lambda x$ of $x$. The scalar has to be the same for all $x$ because if $x$ and $y$ are multiplied by different scalars, then $x+y$ would not be sent to a scalar multiple of itself.

I don't know if one can directly argue that all nonzero elements of $\mathcal{B}(H)'$ are invertible in $\mathcal{B}(H)'$, but it follows.

If $A \in \mathcal{B}(H)$ commutes with every bounded operator, in particular it commutes with all orthogonal projections. That means every subspace of $H$ is an invariant subspace of $A$, in particular all nonzero elements of $H$ are eigenvectors. If $A$ had more than one eigenvalue, the sum of two eigenvectors to different eigenvalues wouldn't be an eigenvector.