Prove that T is diagonalizable if and only if the minimal polynomial of T has no repeated roots.

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

$$(T - \lambda_i I) Q_i(T)x = \eta_i \mu_T(T)x = 0$$

How short this proof can be made depends entirely on your background. Here's a short one that I hope will be accessible to you.

Background facts:

I'm going to assume you are familiar with the notion of a direct sum. In particular, if $T$ acts on $V$ and $V=W\oplus Z$ with $TW\subseteq W$ and $TZ\subseteq Z$, then $T$ splits over the direct sum, and we have $T=T|_W\oplus T|_Z$. Then if $m_X$ denotes the minimal polynomial and $p_X$ denotes the characteristic polynomial of $X$, then whenever we have $A=B\oplus C$, $p_A(t)=p_B(t)p_C(t)$, and $m_A(t)=\newcommand{\lcm}{\operatorname{lcm}}\lcm(m_B(t),m_C(t))$. We'll use these two facts.

Proof:

Now since you're familiar with the generalized eigenspaces, which I will denote $E_{\lambda_i}$, note that $V=\bigoplus_i E_{\lambda_i}$ and $TE_{\lambda_i}\subseteq E_{\lambda_i}$. Thus $m_T(t) = \lcm\{m_{T|_{E_{\lambda_i}}}(t):i\}$. However, $m_{T|_{E_{\lambda_i}}}(t)=(t-\lambda_i)^{n_i}$ where $n_i$ is the least integer such that $(T-\lambda_i)^{n_i}E_{\lambda_i}=0$, so they are all relatively prime. Hence, $m_T(t) = \prod_i (t-\lambda_i)^{n_i}$.

Thus the minimal polynomial of $T$ has no repeated roots if and only if the $n_i$s are 1. Then since $n_i$ by definition is the least integer such that $(T-\lambda_i)^{n_i} E_{\lambda_i}=0$, $n_i=1$ for all $i$ if and only if $T|_{E_{\lambda_i}}=\lambda_i$ for all $i$. And this is the case if and only if the generalized eigenspace is the eigenspace, hence if and only if $T$ is diagonalizable.

Alternative Methods:

Another approach would be to use Jordan canonical form, although I'm not sure whether or not you're familiar with it.