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 $f(x)=(x-c_1)^{d_{1}}\cdots(x-c_k)^{d_k}$.
While I can see no errors in your argument, I would say it achieves this easy result in a rather roundabout way. Note for instance that you introduce an arbitrary annihilating polynomial $f$ but never really use it in the argument; you go straight for the minimal polynomial $m_T$ instead. Which is valid if you know that the minimal polynomial exists: any monic annihilating polynomial whose degree does not exceed that of the minimal polynomial is the minimal polynomial, though there is an smell of circularity when using that in an argument.
I would use the main fact about polynomials in $T$ and eigenvectors instead, namely that $P[T](v)=P[\lambda](v)$ for any polynomial $T$, whenever $T(v)=\lambda v$. For a polynomial $P(T)$ in a diagonalisable $T$ to be $0$, it suffices that it kills all eigenvectors (for it will then be zero on the sum of eigenspaces, which is the whole space), and for this one must have $P[\lambda]=0$ whenever $\lambda$ is an eigenvalue of $T$. Clearly the minimal degree monic polynomial with this property is the product of factors $X-\lambda$ for $\lambda$ running over all distinct eigenvalues.
Best Answer
Note that $TR=RT$, as for any matrix $B$, $$TR(B)=T(BA)=ABA=R(AB)=RT(B).$$
Therefore, $T,R$ can be made diagonal simultaneously in a basis, which means $T-R$ is diagonal in that basis.