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
This theorem is true for arbitrary $V$ (over an arbitrary field $\mathbb{F}$).
We can prove the following
Lemma. If $v_1 + v_2 + \cdots + v_k \in W$ and each of the $v_i$ are eigenvectors of $A$ corresponding to distinct eigenvalues, then each of the $v_i$ lie in $W$.
Proof. Proceed by induction. If $k = 1$ there is nothing to prove. Otherwise, let $w = v_1 + \cdots + v_k$, and $\lambda_i$ be the eigenvalue corresponding to $v_i$. Then:
$$Aw - \lambda_1w = (\lambda_2 - \lambda_1)v_2 + \cdots + (\lambda_k - \lambda_1)v_k \in W.$$
By induction hypothesis, $(\lambda_i - \lambda_1)v_i \in W$, and since the eigenvalues $\lambda_i$ are distinct, $v_i \in W$ for $2 \leq i \leq k$, then we also have $v_1 \in W$. $\quad \square$
Now each $w \in W$ can be written as a finite sum of nonzero eigenvectors of $A$ with distinct eigenvalues, and by the Lemma these eigenvectors lie in $W$. Then we have $W = \bigoplus_{\lambda \in F}(W \cap V_{\lambda})$ as desired (where $V_{\lambda} = \{v \in V\mid Av = \lambda v\}$).