I won't use any other eigenvalues explicitly, so set $\lambda=\lambda_1$ and $r=r_1$ for brevity.
Let $P=(X-\lambda)^r$ and let $Q=f/P$ be the remaining factor of the characteristic polynomial $f$. By assumption (certainly you meant the $\lambda_i$ to be distinct, although this is not said explicitly), $\lambda$ is not a root of$~Q$, and so $P$ and $Q$ are relatively prime in $F[X]$. Therefore there exist Bézout coefficients $S,T\in F[X]$ with $SP+TQ=1$. Since $(PQ)[A]=0$ (by Cayley-Hamilton) it is easy to see that $(SP)[A]$ and $(TQ)[A]$ are projectors on complementary subspaces $U,W$ of $V$. Then $P[A]$ acts invertibly on $U$ and vanishes on $W$, so $W=\ker(P[A])=\ker((A-\lambda I)^r)$, and $U$ is the image of $P[A]$, in particular $\operatorname{rank}(P[A])=\dim(U)$.
Also the characteristic polynomial $f$ of $A$ is the product of the characteristic polynomials of the restrictions of$~A$ to $U$ and $W$, the first of which does not have $\lambda$ as a root (as $P[A]$ acts invertibly on $U$) while the latter is a power of $X-\lambda$ (since $(X-\lambda)^r$ is an annihilating polynomial). But then the characteristic polynomial of the restriction of$~A$ to$~W$ is $(X-\lambda)^r$, and $\dim(W)$ is its degree$~r$, and $\dim(U)=\dim(V)-r$, completing the proof.
Note that my second paragraph only depends of $PQ$ being an annihilating polynomial with $P$ grouping all its factors $X-\lambda$; in particular it could have been the minimal polynomial, and the exponent could have been different from $r$. This is in fact a bit confusing in a proof of something that eventually says something about$~r$. Eventually $r$ is gotten as $\dim(W)$, not as $\deg(P)$.