Quantization of angular momentum: is Dirac's proof wrong?

I don't know the mentioned book.
However, about Step 2, $J_x, J_y, J_z$ are generators of an unitary representation of $SO(3)$ that is a compact Lie group. By Peter-Weyl theorem all irreducible unitary representation of a compact Lie group are finite-dimentional. So $J_z, J^2$ are made up of self-adjoint finite-dimensional operators and as such they admits at least an eigenvector.

Edit

Let $J_x, J_y, J_z$ be an irreducible representation of the above commutation rules.
By Schur's lemma, $[J^2, J_i] = 0$ for each $i\in\{x, y, z\}$, implies $$ J^2 = \beta \mathbb I $$ $J_z$ is a finite-dimensional self adjoint operator, hence diagonalizable.
Let's take an eigenvector, $\lvert m \rangle$, of $J_z$, with repeated applications of $J_-$ and $J_+$ we can construct a sequence of eigenvectors $$ s := \{\lvert m - h \rangle, \dotsc, \lvert m - 1\rangle, \lvert m \rangle, \lvert m + 1 \rangle, \dotsc,\lvert m + k \rangle \} $$ Above sequence is finite because its elements are independent vectors of the finite-dimensional representation space $V$.
Let's call $S$ the subspace spanned by vectors in $s$. The equalities $$ J_x = \frac 1 2 (J_+ + J_-)\\ J_y = \frac 1 {2i} (J_+ - J_-) $$ show that $S$ is an invariant space for $J_x, J_y, J_z$. Since the representation is irreducible $S$ must coincide with $V$ and no other (indipendent) eigenvectors of $J_z$ can exist.