Instead of just "defining" $U$ as "consisting of infinite many segments", you should define it explicitly -- for example as all of $X$ except for the horizontal base. (Actually, since $U$ has to be a neighborhood of $x_0$, it has to contain (parts of) infinitely many teeth of the comb, so your choice is not a choice at all).
The last step is a little roundabout. It would be simpler to note that any neighborhood $V\subseteq U$ of $x_0$ contains points with a different first coordinate, and such a point has no path to $x_0$ in $U$.
You must distinguish between a deformation retract and a strong deformation retract. A retraction $r : Z \to A$ to a subset $ A \subset Z$ is a deformation retraction if there exists a homotopy $H : Z \times [0,1] \to R$ such that $H(z,0) = z$ and $H(z,1) = r(z)$. If exists a homotopy such that in addition $H(a,t) = a$ for all $a \in A, t \in [0,1]$, then $r$ is called a strong deformation retraction. See https://en.wikipedia.org/wiki/Retract.
A strong deformation retraction to a point $p_0 \in Y = [0,1] \times \{0\}$ can be composed by the obvious strong deformation retraction of $X$ to $Y$ and the obvious strong deformation retraction of $Y$ to $p_0$.
Why doesn't this work for $p_0 \in U = X \setminus Y$? Note that $U$ is open in $X$ and has infinitely many path components $U_r = \{r\} \times (0,1-r]$, $r \in [0,1) \cap \mathbb Q$. Since $\mathbb Q$ is dense in $\mathbb R$, any open subset of $X$ intersects infinitely many of the $U_r$.
Now assume we have a strong deformation retraction $r : X \to \{p_0\}$. There is a homotopy $H : X \times [0,1] \to X$ such that $H(x,0) = x$, $H(x,1) = p_0$ and $H(p_0,t) = p_0$. Since $U$ is an open neigborhood of $p_0$ and $H(\{p_0 \} \times [0,1]) = \{p_0\} \subset U$, there exist an open neighborhood $V$ of $p_0$ such that $H(V \times [0,1]) \subset U$. We have $p_0 \in U_{r_0}$ for some $r_0 \in [0,1)$. For each $p \in V$ the set $H(\{p\} \times [0,1])$ is a path connected subset of $U$ and contains $H(p,1) = p_0$, thus it must be contained in the path component $U_{r_0}$ of $p_0$ in $U$. But then also $p = H(p,0) \in U_{r_0}$. Therefore $V \subset U_{r_0}$. This is a contradiction because any open $V$ must intersect infinitely many $U_r$.
What does it mean that $\{p_0\}$ is a deformation retract of $X$? It means nothing else than that the identity on $X$ is homotopic to the constant map with value $p_0$. But we already know that $X$ is contractible, hence all constant maps to $X$ are homotopic, and this proves that $\{p_0\}$ is a deformation retract of $X$ for any $p_0 \in X$.
Best Answer
This depends on what you mean by being "specifically constructed for this purpose". A nice example (that I'm sure inspired the example given by Hatcher) is the comb space $$X=[0,1]\times\{0\}\cup\{0\}\times [0,1]\cup\bigcup_{n=1}^\infty \left(\left\{\frac{1}{n}\right\}\times [0,1]\right)$$
Any connected subset of the leftmost tooth $\{0\}\times[0,1]$ is a deformation retract of $X$, since $X$ contracts to the point $(0,0)$. But most of the time these won't be strong deformation retracts; in fact the only subset of that tooth that is a strong deformation retract is the point $(0,0)$.