I proceeded proving the statement as it follows:
$f:V\to V$ being diagonalizable implies that there exist $\mathscr{B} = \left( v_1,\dots ,v_n \right)$, a base of eigenvectors for $V$; the fact that for a fixed $i=1, \dots, n$ there exist $V_{\lambda_i}$ such that $dimV_{\lambda_i} \gt 1$, implies that $ker\left( f - \lambda_i I\right)=\langle v_1, \dots, v_l\rangle$, with $1 \lt l \le n$. If for absurd there is a cyclic vector $v\in V$, there is $\left(v, f(v), \dots, f^{n-1}(v)\right)$ a collection of L.I. vectors, which means that the Vandermonde matrix is invertible and it's determinant is different than $0$. Now the existence of $V_{\lambda_i}$ with dimension grater than $1$ implies that $\mu_g\left( \lambda_j\right)\gt 1$, which further implies $\mu_a\left( \lambda_j\right)\gt 1$; so there are at least two equivalent eigenvalues in $\left( \lambda_1, \dots, \lambda_n\right)$, and since the determinant of a matrix with two equal columns is $0$ this is absurd. Therefore the thesis.
It seems correct to me but also kind of mechanical; if anyone feels he has a smoother way to prove it I'd be happy to read about it!
Your proof looks correct. Here's a shorter proof.
Let $\lambda_1,\dots,\lambda_k$ (with $k < n$) denote the eigenvalues of $f$ (without repeats). Let $p$ denote the polynomial $p(x) = (x-\lambda_1)\cdots(x-\lambda_k)$, which is of degree $k$. The fact that $f$ is diagonalizable implies that $p(f) = 0$. Thus, there exist cofficients $a_0,\dots,a_k$ such that $a_0 I + a_1 f + \cdots + a_kf^k = 0$. Thus, for every vector $v \in V$, $a_0 v + a_1 f(v) + \cdots + a_kf^k(v) = 0$. Thus, for any vector $v$, the set $$ \{v,f(v),\dots,f^k(v)\} \subseteq \{v,f(v),\dots,f^{n-1}(v)\} $$ is linearly dependent. Thus, no vector $v \in V$ can be a cyclic vector.