Proof verification: polynomials $\mathbb R[X]$ are a vector space that is not isomorphic to its dual

Some comments. The dual space $V^{\ast}$ can be identified with real formal power series $\mathbb{R}[[Y]]$; the identification sends a formal power series $\sum a_i Y^i$ to the linear functional

$$\mathbb{R}[X] \ni \sum b_i X^i \mapsto \sum b_i a_i \in \mathbb{R}.$$

With respect to this identification the evaluation homomorphisms correspond to the formal power series $\frac{1}{1 - rY}$. There are many ways to show that these formal power series are linearly independent; maybe the shortest one is to notice that, interpreting them as genuine functions of a real-valued variable $Y$, any nontrivial linear combination of them has a pole somewhere (and hence isn't identically zero).

Any other method of writing down an uncountable family of linearly independent formal power series also constitutes a proof that $V^{\ast}$ is not isomorphic to $V$; another very similar family is the family $\exp (r Y)$. A cute way to show that these power series are linearly independent is to repeatedly differentiate any nontrivial linear combination of them; you can also observe that, again interpreting them as genuine functions of a real-valued variable $Y$, in any nontrivial linear combination one term has the uniquely largest growth rate as $Y \to \infty$ and hence eventually dominates the others.

This is indeed correct. Note that you don't need to talk about a basis of $V^*$ (something whose existence depends on the Axiom of Choice): simply the fact that $V^*$ has an uncountable linearly independent set, while $V$ does not, establishes that they can't be isomorphic.