Clearly $T$ is diagonalizable if and only if we can decompose $V$ into a direct sum of eigenspaces $$V = \ker (T-\lambda_1I) \dot+ \ker(T - \lambda_2 I) \dot+ \cdots \dot+\ker(T - \lambda_k I)$$
since we can then take a basis of the form $$(\text{basis for }T-\lambda_1I, \text{basis for }T-\lambda_2I, \ldots, \text{basis for }T-\lambda_nI)$$
which yields a diagonal matrix representation of $T$.
You have already handled the direction ($T$ is diagonalizable $\implies$ minimal polynomial has no repeated roots).
Conversely, assume that the minimal polynomial $\mu_T$ has no repeated roots. Note that the above sum is direct:
$$x \in \ker(T - \lambda_i I) \cap \ker(T - \lambda_j I) \implies \lambda_ix = Tx = \lambda_jx \implies i = j \text{ or } x = 0$$
It remains to prove that every $x$ can be written in the form $x = x_1 + \cdots + x_n$ with $x_i \in \ker(T - \lambda_iI)$.
Using the partial fraction decomposition we obtain:
$$\frac1{\mu_T(x)} = \frac1{(x-\lambda_1)\cdots(x-\lambda_k)} = \sum_{i=1}^k \frac{\eta_i}{(x-\lambda_i)}$$
for some scalars $\eta_i$.
Define $$Q_i(x) = \frac{\eta_i \mu_T(x)}{x - \lambda_i}$$ so that $\sum_{i=1}^n Q_i = 1$ and $(x-\lambda_i)Q_i(x) = \eta_i \mu_T(x)$.
Finally, notice that the desired decomposition is given by $$x = Q_1(T)x + Q_2(T)x + \cdots + Q_k(T)x$$
with $Q_i(T) x \in \ker (T - \lambda_i I)$ since
$$(T - \lambda_i I) Q_i(T)x = \eta_i \mu_T(T)x = 0$$
Best Answer
As requested, I'll answer the question in the comments:
If $A$ is diagonalisable, then $A$ is similar to a diagonal matrix. Diagonal matrices are automatically in Jordan Canonical Form; they are block diagonal matrices with $1 \times 1$ Jordan cells. This means that one of the Jordan Canonical Forms (they are not, strictly speaking, unique) is a diagonal matrix.
Now, as I say, the JCF is not unique (typically). Of course, there is some rigidity here though: the JCF is unique up to a permutation of the Jordan Blocks along the diagonal. So, if a Jordan block appears in one Jordan Canonical Form of a matrix, then it will appear in every other JCF of the matrix, just in various different positions.
So, to sum up, if $A$, a complex $n \times n$ matrix, is diagonalisable, then $A$ is similar to a diagonal matrix $D$. $D$ is a matrix already in Jordan normal form, as each of the $1 \times 1$ submatrices along the diagonal are (trivially) Jordan Blocks. By the uniqueness result alluded to above, this means that every JCF of the matrix has the same $1 \times 1$ Jordan Blocks.