Prove $D \in \mathcal{L}(\mathcal{P}(\mathbf{R}),\mathcal{P}(\mathbf{R})) : \text{deg}(D(p)) = \text{deg}(p) - 1$ is surjective

I'm not sure if this is the most efficient solution for this problem, but I'll give it a try.

We can first show that the subspace $\mathbb{R}_{m}[x]$ of polynomials of degree at most $m$ is contained in the image of $D$, $\operatorname{im } D$, for any $m \geq 0$. This will imply that the image of $D$ has polynomials of all degrees, so it should indeed be all of $\mathbb{R}[x]$. For that purpose, the following result will come in handy:

$\textbf{Lemma}$. Let $p_{0}, \ldots, p_{m}$ be $m + 1$ polynomials such that $\deg p_{i} = i$ for $i = 0, \ldots, m$. Then $p_{0}, \ldots, p_{m}$ is a basis for $\mathbb{R}_{m}[x]$.

$\textit{Proof.}$ I can expand on this if you wish.

Let $m \geq 0$ be arbitrary and consider the $m + 1$ nonconstant polynomials $x, \ldots, x^{m+1}$. Now let's take a look at their values under $D$:

$$ D(x), \ldots, D(x^{m+1}) .$$

By the hypothesis of the problem, we know that these polynomials have degrees from $0$ to $m$, so they form a basis for $\mathbb{R}_{m}[x]$. In particular:

$$ \mathbb{R}_{m}[x] = \operatorname{span}(D(x), \ldots, D(x^{m+1})) $$

Notice that $D(x), \ldots, D(x^{m + 1})$ are polynomials in $\operatorname{im }D$, which is a subspace of $\mathbb{R}[x]$. A fundamental property of $\operatorname{span}(D(x), \ldots, D(x^{m+1}))$ is that it is the smallest subspace containing $D(x), \ldots, D(x^{m+1})$. We can therefore deduce that

$$ \mathbb{R}_{m}[x] = \operatorname{span}(D(x), \ldots, D(x^{m+1})) \subseteq \operatorname{im } D .$$

Now consider an arbitrary polynomial $p(x) = a_{0} + a_{1}x + \ldots + a_{m}x^{m}$. Then $p \in \mathbb{R}_{m}[x]$, so $p$ must be in the image of $D$ as well. Since $p$ was an arbitrary, we can conclude that $\mathbb{R}[x] \subseteq \operatorname{im }D$, so $ \mathbb{R}[x] = \operatorname{im }D $ and $D$ is surjective.