Firstly, we assume $\mathscr{dim}~V=n$.
From the definition of Hoffman’s Linear Algebra, $T$ is diagonalizable if there is a basis for $V$ each vector of which is a characteristic vector of $T$.
The author has supposed $c_1,c_2,\cdots,c_k$ are all the distinct characteristic values of $T$.
So there exists $({\alpha_1,\alpha_2,\cdots,\alpha_n})$ which are basis.
Here is a step the author omits that we can change the positions of $\alpha_p$ and $\alpha_q$ if necessary such that $1\le i\le d_1$, $\alpha_i$ is the eigenvector of $c_1$ , $d_{1}+1\le j\le d_2+d_1$, $\alpha_j$ is the eigenvector of $c_2$ and so on.
Thus
$$T({\alpha_1,\alpha_2,\cdots,\alpha_n})=({\alpha_1,\alpha_2,\cdots,\alpha_n})\left( \begin{matrix}
c_1I_1& & & \\
& c_2I_2& & \\
& & \ddots& \\
& & & c_kI_k\\
\end{matrix} \right).$$
We use the following property:
If the matrix of $T$ under a basis of $V$ has the form $$\left( \begin{matrix}
A_1& *\\
0& A_2\\
\end{matrix} \right) ,$$we define the characteristic polynomial for $T$ is $f(x)$, the characteristic polynomial for $A_1$ is $f_1(x)$ and the characteristic polynomial for $A_2$ is $f_2(x)$, then $f(x)=f_1(x)f_2(x)$.
So we get the characteristic polynomial for $T$ is $f(x)=(x-c_1)^{d_{1}}\cdots(x-c_k)^{d_k}$.
I find both your proof and Hoffman's needlessly complicated. Also the terminology is somewhat confusing me (I've difficulty reconciling "let $p_\alpha$ be the $T$-annihilator of $\alpha$" with "$p_\alpha q$ is in the $T$-annihilator of $\alpha$". Let me call $p_\alpha$ the $\alpha$-minimal polynomial of $T$, and any polynomial in the ideal $S_T(\alpha;\{0\})$ it generates a $\alpha$-annihilating polynomial of $T$.
This all becomes easy if you start like this. The infinite sequence of vectors $\alpha,T(\alpha),T^2(\alpha),\ldots$ cannot be linearly independent (in finite dimension) so there is some $k$ such that $\def\B{\mathcal{B}}\B=[\alpha,T(\alpha),T^2(\alpha),\ldots,T^{k-1}(\alpha)]$ is linearly independent, but adding $T^k(\alpha)$ makes it linearly dependent (we have $k>0$ since $\alpha\neq0$). Now the span $Z$ of $\B$ is $T$-stable (as $T$ maps all vectors of $\B$ into $Z$) and $\B$ is a basis of $V$ (nothing left to prove here). Clearly all powers $T^i(\alpha)$ remain inside the $T$-stable subspace $Z$, so $Z$ is the cyclic subspace $Z(\alpha;T)$.
The coordinates of$~T^k(\alpha)$ with respect to the basis$~\B$ directly determine the $\alpha$-minimal polynomial $p_\alpha$ of $T$, which is of degree$~k$.
What remains to be shown is part of (iii): that $p_\alpha$ is an annihilating polynomial of the restriction$~U$ of$~T$ to$~Z$ (knowing that it is $\alpha$-minimal, it will then also be minimal on$~U$). But $\ker(p_\alpha[T])$ contains the vector $\alpha$ and is $T$-stable (since $p_\alpha[T]$ commutes with $T$, see the last line of Hoffman's proof), so it contains the cyclic subspace $Z$ generated by$~\alpha$; this proves the point.
Just to add a cross reference, see also this question.
Best Answer
Indeed, it is well known, and quite obvious (look what $Q[M](e_1)$ is when $\deg(Q)\leq{k}$), that the minimal polynomial of a companion matrix$~M$ of $P$ is $P$ itself; in particular its degree equals the dimension of the vector space acted upon. Since the minimal polynomial is a property of a linear operator, it must be the same for any matrix of the operator, so by contrapositive, if the degree of the minimal polynomial of a linear operator is strictly less than the dimension of the space, it cannot have a companion matrix on any base.
The converse is also true: if the degree of the minimal polynomial of a linear operator equals the dimension of the space (so that the minimal polynomial itself coincides with the characteristic polynomial), then there exists a cyclic vector and so a basis for which the matrix it the companion matrix of the minimal polynomial. There are several ways to characterise this situation.