I am reading "Linear Algebra Done Right 3rd Edition" by Sheldon Axler.
There is the following example on p.138.
5.15 Example (b)
Define an operator $T \in \mathcal{L}(\mathbb{F}^2)$ by $T(x, y) = (y, 0)$.
Let $U = \{(x, 0) : x \in \mathbb{F}\}$.
Show that there does not exist a subspace $W$ of $\mathbb{F}^2$ that is invariant under $T$ and such that $\mathbb{F}^2 = U \oplus W$.Solution
Suppose $W$ is a subspace of $V$ such that $\mathbb{F}^2 = U \oplus W$. Because $\dim \mathbb{F}^2 = 2$ and $\dim U = 1$, we have $\dim W = 1$. If $W$ were invariant under $T$, then each nonzero vector in $W$ would be an eigenvector of $T$. However, it is easy to see that $0$ is the only eigenvalue of $T$ and that all eigenvectors of $T$ are in $U$. Thus $W$ is not invariant under $T$.
Axler concluded that $W$ is not invariant under $T$.
I think there is a very small gap.
So, I tried to fill the gap:
We conclude that $W \subset U$.
And then, $\mathbb{F}^2$ is not a direct sum of $U$ and $W$.
And we get a contradiction.
So, $W$ was not invariant under $T$.
Is my argument correct or not?