I am reading "Introduction to Linear Algebra" (in Japanese) by Kazuo Matsuzaka.
There is the following problem in this book:
Let $V$ be a finite-dimensional vector space.
Let $F$ be a linear map on $V$.
Then, $$\text{Ker } F \subset \text{Ker } F^2 \subset \text{Ker } F^3 \subset \cdots$$ and $\dim(\text{Ker } F^k) \leq \dim V$.
So, there exists a positive integer $q$ such that $$\text{Ker } F^q = \text{Ker } F^{q+1}.$$Prove the following facts:
(a) $\text{Ker } F^k = \text{Ker } F^q$ for any integer $k \ge q$.
(b) $V = \text{Ker } F^q \oplus \text{Im } F^q$ and $F$ is nilpotent on $\text{Ker } F^q$ and $F$ is non-singular on $\text{Im } F^q$.
(c) Let $U$ be an $F$-invariant subspace such that $F$ is nilpotent on $U$. Then $U \subset \text{Ker } F^q$.
(d) Let $W$ be an $F$-invariant subspace such that $F$ is non-singular on $W$. Then $W \subset \text{Im } F^q$.
I think the following (c') is more general than (c) above.
I think $U$ doesn't need to be an $F$-invariant subspace.
The author didn't write (c') instead of (c).
Why?
(c') Let $U$ be a subspace such that $F : U \to V$ is nilpotent. Then $U \subset \text{Ker } F^q$.
Proof:
$F^l(u) = 0$ for any $u \in U$ for some $l$.
So, $U \subset \text{Ker } F^l$.
If $l \geq q$, then $\text{Ker } F^l = \text{Ker } F^q$ by (a). So, $U \subset \text{Ker } F^q$.
If $l < q$, then $U \subset \text{Ker } F^l \subset \text{Ker } F^q$.
Your proof of (c') is almost correct. Let $u \in U$, we show that $u \in \ker(F^q)$ :
Let $l$ be such that $u \in \ker(F^l)$.
if $l \leqslant q$, then $u \in \ker(F^l) \subseteq \ker(F^q)$.
if $l > q$ then by (a) we get $u \in \ker(F^l) = \ker(F^q)$.