Contractibility is Not Equivalent to One-Point Homotopy Type (9.58.8)

James Munkres Topology, chapter 9.58, exercise 8:

Find a space $X$ and a point $x_0∈X$ such that the inclusion $\{x_0\}→X$ is a homotopy equivalence, but $\{x_0\}$ is not a deformation retract of $X$.

Proof: Let $X$ be the subset of $ℝ^2$ consisting of those lines $(1/n)×I$ for $n∈ℕ$, as well as $0×I$ and $I×0$, and let $x_0=(0,1)$. Then the map $$F : X×I→X$$ $$F((x,y),t)=\left\{ \begin{array} \{ (x,(1-3)y) & t∈[0,1/3] \\ ((2-3t)x,0) & t∈[1/3,2/3] \\ (0,3t-2) & t∈[2/3,1] \end{array} \right.$$ is a homotopy between the identity on $X$ and the constant map onto $x_0$, so that $X$ is contractible. But suppose $\{x_0\}$ is a deformation retract of $X$ via the map $F : X×I→X$, i.e. a homotopy between the two mentioned above such that $F(x_0×I)=\{x_0\}$. For each $n$ let $x_n=(1/n)×1$; then since each $ρ_n : I→X$ given by $ρ_n(t)=F(x_n,t)$ is a path from $x_n$ to $x_0$, let $t_n∈I$ be such that $π_2(F(x_n,t_n))=0$. Since $I$ is compact, let $t_{n_i}→α$ be a convergent subsequence. Then $(x_{n_i},t_{n_i})→(x_0,α)$ yet $F(x_{n_i},t_{n_i}) \not → x_0$ since every term of this sequence has second coordinate $0$.$~\square$