This is how the proof starts:
From here, the author introduces another operator to get around the issue that $W$ might not be invariant under $T$ and follows up with some more algebra. But I think that $W$ is actually invariant, if we add the assumption that $T$ is invertible (if $T$ isn't invertible, then $0$ is an eigenvalue and we're done). It seemed odd to me that Axler might have missed that, so I wanted to verify my reasoning:
Claim :
Suppose $T$ is invertible. Then $W$ is invariant under $T$.
Proof :
If $T$ is invertible, then it's injective. So $T|_U$ is injective, and therefore invertible (1).
Now suppose for contradiction that there is some $w \in W, w \neq 0$ such that $Tw \notin W$. Since $V = U \oplus W$, then $U \cap W = \{0\} \ $. It follows that $\ Tw \in U$. From (1), it follows that $w\in U$. Contradiction.
If I've got something wrong, I'm guessing it's either the very first or very last step; but it seems good to me, which is weird.
Best Answer
It is not necessarily the case that $W$ is invariant, since $W$ is any subspace for which $V = U \oplus W$. As an example, consider the transformation $T$ on $\Bbb C^2$ associated with the matrix $$ A = \pmatrix{1&2\\0&-1}. $$ The "$x$-axis" $U = \{(t,0): t \in \Bbb C\}$ is an invariant subspace. The "$y$-axis" $W = \{(0,t): t \in \Bbb C\}$ is complementary to $U$, but it is not an invariant subspace. For example, $(0,1) \in W$, but $T(0,1)\notin W$.
This transformation is invertible, however. In fact, $T$ is its own inverse.