Let $T:V \to V$ be a linear transformation (where $V$ is a vector space over a field $K$).
Suppose the Characteristic Polynomial of $T$ is $C_T(x) = {(x-\lambda_1)}^{r_1}{(x-\lambda_2)}^{r_2}\dots{(x-\lambda_k)}^{r_k}$ (where $\lambda_i$ are all distinct eigenvalues and $r_i$ are the algebraic multiplicities).
Then, will the Minimal Polynomial of $T$ be $M_T(x)=(x-\lambda_1)(x-\lambda_2)\dots(x-\lambda_k)$?
If it is not always true, can you provide a counter example?
No, the minimal polynomial of a linear operator does not always have simple roots (even when the characteristic polynomial is split, your assumption which implies that the minimal polynomial is also split). In fact there is a theorem that says that the operator is diagonalisable if and only if its minimal polynomial is split and has simple roots; this characterises precisely the cases where your claim fails.
And any monic polynomial $P$ can be the minimal polynomial of a linear operator; it suffices to take the operator whose matrix (on some basis) is the companion matrix of $P$ (or the transpose of that companion matrix; either will do fine).