In my opinion, the solution is not totally convincing. It only shows that there is one representation of the set $S$ with infinitely many inequalities. It does not show that there might be another representation with finitely many inequalities.
Of course, for the quarter of the circle $S$, this might be obvious, but it is not mentioned in the solution at all. If this would be an answer of a student in a test or homework, I would not give all possible points.
To give another example:
$$R = \{ x \in \mathbb R \mid x \, t \le 1 \; \forall t \in [-1,1]\}.$$
Again, $R$ is defined by infinitely many hyperplanes, but is a polyhedron since $R = [-1,1]$.
Here is a very tedious answer. I imagine there is a much slicker solution, but it escapes me.
Note that we can write $S = \{ x| (1,t,t^2)^T x\le {11 \over 10}, t \in (0,1] \}$. Since $S$ is the intersection of closed halfplanes it is convex and closed.
Let $S_0 = \{ x \in S | x_1 = 0 \}$ and note that if $S$ was polyhedral then $S_0$ would be too. Hence it suffices to show that $S_0$ is not polyhedral.
Just to reduce noise (I am switching use of $x$ here), let $S_0' = \{ (x,y)| tx+t^2 y \le 1.1, t \in (0,1]\} $.
Note that if $(x,y) \in S_0'$ then $(x-h,y) \in S_0'$
for all $h \ge 0$. Furthermore there is some $l>0$ such that $(x+l,y) \notin S_0'$. In addition, for any $y$ there is some $x$ such that $(x,y) \in S_0'$.
Hence we can characterise $S_o'$ by computing
$f(y) = \max_{(x,y) \in S_0'} x$ (the $\max$ exists because $s_0'$ is closed) and write
$S_0' = \{(x,y) | x \le f(y) \}$.
We can write $tx+t^2y \le 1.1$ as $x \le {1.1 \over t} - ty$
and so we see that $f(y) = \inf_{t \in (0,1]} ({1.1 \over t}-t y)$.
If $y \ge 0$ then $t \mapsto {1.1 \over t}-t y$ is decreasing and so $f(y) = 1.1-y$.
If $y < 0$ then $t \mapsto {1.1 \over t}-t y$ is unimodal on $(0,\infty)$ and has a unique $\min$ in $t^* = \sqrt{1.1 \over -y}$.
In particular, for $y \ge - 1.1$, $f(y) = 1.1-y$ and for
$y < -1.1$ we have
$f(y) = 2 \sqrt{-1.1y}$.
It is straightforward to show from this that $S_0'$ is not polyhedral.
Best Answer
To make my comment into an answer:$$P = \{ x | Ax \leq b, Cx =d\}=\{x | \begin{pmatrix} A\\ C \\ -C\end{pmatrix}x\leq\begin{pmatrix}b\\d\\-d \end{pmatrix}\}$$So the $Cx=d$ in your definition is just describing more inequalities. Whether they are redundant or not, depends on the polyhedron.