Show that $1, (x-5)^2, (x-5)^3$ is a basis of the subspace $U$ of $\mathcal P_3(\Bbb R)$

I think what Axler is saying is that $\dim U$ cannot be equal to 4 because $U$ is not all of $\mathcal P_3(\mathbb R)$. It is easy to see that they are not equal by just finding a polynomial whose derivative at $x=5$ is not 0. (For example, try $p(x)=x$.) If you take this polynomial and add it to a basis for $U$, you should still have a linear independent set because it is not a linear combination of the basis elements of $U$. But if there are already 4 elements in a basis for $U$ then we would end up with 5 independent vectors in $\mathcal P_3(\mathbb R)$, which is impossible as the dimension of this space is 4.

So I guess Axler is saying, whatever the dimension of $U$ is, you can add vectors to a basis for $U$ until you get a basis for $\mathcal P_3(\mathbb R)$, and you would have to add at least one vector.

$\dim P_3(\Bbb R) = 4$ because $\{1, x, x^2, x^3\}$ is a basis of $P_3(\Bbb R)$.

Clearly $p'(5) = 0$ does not hold for all $p \in P_3(\Bbb R)$.

So your $U$ is a subspace of $P_3(\Bbb R)$ which does not contain all the vectors of $P_3(\Bbb R)$.

Therefore, its dimension must be lower.

Abstract-Algebra Argument

The linear map $\phi:\mathcal{P}_3(\mathbb{R})\to\mathbb{R}$ sending $f\mapsto f'(5)$ is surjective with kernel $U$. Therefore, the Rank-Nullity Theorem gives $$\dim_\mathbb{R}(U)+1=\dim_\mathbb{R}(U)+\dim_\mathbb{R}(\mathbb{R})=\dim_\mathbb{R}\big(\mathcal{P}_3(\mathbb{R})\big)=4\,.$$