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
$$(T - \lambda_i I) Q_i(T)x = \eta_i \mu_T(T)x = 0$$
A typical 2 x 2 non-diagonalizable matrix is
$$\pmatrix{
1 & 1 \\ 0 & 1}
$$
Its characteristic polynomial has one double-root, but its minimal polynomial is also $(x-1)^2$, which makes it different from the identity, whose char. poly has a double root, but whose minimal polyonomial is $(x-1)$.
What your prof. said was correct, but you negated it incorrectly. :)
By the way, I applaud your questioning this. Asking questions like this, even ones that seem stupid, is part of how you learn to recognize certain classes of errors and learn not to make them again. Go, you!
Best Answer
As pointed out in the comments, this is not true.
More specifically, the roots of the minimal polynomial $\mu_A(t)$ of a matrix $A$ are eigenvalues of $A$, but if the minimal polynomial contains irreducible non-linear factors, then $A$ is not diagonalizable. Vice versa, if $A$ is diagonalizable, then $\mu_A$ can be linearly decomposed.
An example can be a matrix $A\neq 0$ with minimal polynomial $\mu(t)=t^2(t-1)$, such as $$A=\begin{pmatrix}0& 1&0\\0 &0&0\\0&0&1\end{pmatrix}.$$ The minimal polynomial of such $A$ is $\mu_A(t)=t^2(t-1)$. The eigenvalues of $A$ are $0$ and $1$. However, the nonlinear factor $t^2$ in the minimal polynomial means that $A$ is not diagonalisable, even though the minimal polynomial does have distinct roots.